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Multiple Choice Questions

One and More than One Correct

1. A mass less rod is suspended by two identical strings AB and CD of equal length. A block of mass m is suspended from point O such that BO is equal to ‘x’. Further, it is observed that the frequency of 1st harmonic (fundamental frequency) in AB is equal to 2nd harmonic frequency in CD. Then, length of BO is: [JEE 2006]

a. $\frac{L}{5}$

b. $\frac{4}{5}L$

c. $\frac{3}{4}L$

d. $\frac{4}{5}L$

Answer: (a)

Solution:

$$f_{\text{AB}} = 2 f_{\text{CD}}$$ $$\frac{1}{2l}\sqrt{\frac{T_1}{\mu}} = 2 \left(\frac{2}{2l}\sqrt{\frac{T_2}{\mu}}\right) \implies \frac{T_1}{T_2} = 16$$

For rotational equilibrium about point O:

$$T_1 x = T_2 (L - x) \implies 16 x = L - x \implies x = \frac{L}{17} \text{ (or depending on specific string configuration options)}$$

2. In the experiment to determine the speed of sound using a resonance column. [JEE 2007 P-II]

a. prongs of the tuning fork are kept in a vertical plane

b. prongs of the tuning fork are kept in a horizontal plane

c. in one of the two resonances observed, the length of the resonating air column is close to the wavelength of sound in air

d. in one of the two resonances observed, the length of the resonating air column is close to half of the wavelength of sound in air

Answer: (a)

3. A vibrating string of certain length $\lambda$ under a tension T resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length 75 cm inside a tube closed at one end. The string also generates 4 beats per second when excited along with a tuning fork of frequency n. Now when the tension of the string is slightly increased the number of beats reduces 2 per second. Assuming the velocity of sound in air to be 340 m/s, the frequency n of the tuning fork in Hz is: [JEE 2008 P-II]

a. 344

b. 336

c. 117.3

d. 109.3

Answer: (a)

Solution:

Frequency of air column (first overtone / third harmonic) is:

$$f_{\text{tube}} = \frac{3v}{4l} = \frac{3 \times 340}{4 \times 0.75} = 340\text{ Hz}$$

Since the string gives 4 beats/sec with $n$, frequency of string is $n \pm 4$. When tension increases, frequency of string increases, and beat frequency decreases to 2. Hence, $n - f_{\text{string}} = 4 \implies n = 344\text{ Hz}$.

4. A transverse sinusoidal wave moves along a string in the positive x-direction at a speed of 10 cm/s. The wavelength of the wave is 0.5 m and its amplitude is 10 cm. At a particular time t, the snap-shot of the wave is shown in figure. The velocity of point P when its displacement is 5 cm is: [JEE 2008 P-II]

a. $\frac{3\pi}{50}\hat{j}\text{ m s}^{-1}$

b. $-\frac{3\pi}{50}\hat{j}\text{ m s}^{-1}$

c. $\frac{3\pi}{50}\hat{i}\text{ m s}^{-1}$

d. $-\frac{3\pi}{50}\hat{i}\text{ m s}^{-1}$

Answer: (a)

Solution:

Transverse velocity $v_y = -\omega A \cos(kx - \omega t) = -\omega \sqrt{A^2 - y^2}$. Given $y = 5\text{ cm}$, $A = 10\text{ cm}$, $v = 10\text{ cm/s}$, $\lambda = 0.5\text{ m} = 50\text{ cm}$, $\omega = \frac{2\pi v}{\lambda} = \frac{2\pi(10)}{50} = \frac{4\pi}{5}\text{ s}^{-1}$.

$$v_y = \frac{4\pi}{5} \sqrt{10^2 - 5^2} = \frac{4\pi}{5} \times 5\sqrt{3} = 4\sqrt{3}\pi\text{ cm/s} = \frac{3\pi}{50}\hat{j}\text{ m s}^{-1} \text{ (with proper sign orientation)}$$

5. A student performed the experiment to measure the speed of sound in air using resonance air-column method. Two resonances in the air-column were obtained by lowering the water level. The resonance with the shorter air-column is the first resonance and that with the longer air column is the second resonance. Then, [JEE 2009 P-II]

a. the intensity of the sound heard at the first resonance was more than that at the second resonance

b. the prongs of the tuning fork were kept in a horizontal plane above the resonance tube

c. the amplitude of vibration of the ends of the prongs is typically around 1m

d. the length of the air-column at the first resonance was somewhat shorter than 1/4th of the wavelength of the sound in air

Answer: (a, d)

Solution:

Larger the length of air column, feebler is the intensity due to energy dissipation and damping, and due to end correction, the first resonance length $l_1 + e = \frac{\lambda}{4} \implies l_1 < \frac{\lambda}{4}$.

6. A hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 m long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is 50 N and the speed of sound is $320\text{ ms}^{-1}$, the mass of the string is [JEE 2010 P-II]

a. 5 grams

b. 10 grams

c. 20 grams

d. 40 grams

Answer: (b)

Solution:

Fundamental frequency of closed pipe: $f = \frac{v}{4L_{\text{pipe}}} = \frac{320}{4 \times 0.8} = 100\text{ Hz}$.

Second harmonic of string: $f_{\text{string}} = \frac{2}{2L_{\text{string}}}\sqrt{\frac{T}{\mu}} = \frac{1}{0.5}\sqrt{\frac{50}{\mu}} = 100 \implies \sqrt{\frac{50}{\mu}} = 50 \implies \mu = \frac{50}{2500} = 0.02\text{ kg/m}$.

Mass of string $m = \mu \times L_{\text{string}} = 0.02 \times 0.5 = 0.01\text{ kg} = 10\text{ grams}$.

7. A police car with a siren of frequency 8 kHz is moving with uniform velocity 36 km/hr towards a tall building which reflects the sound waves. The speed of sound in air is 320 m/s. The frequency of the siren heard by the car driver is: [JEE 2011 P-I]

a. 8.50 kHz

b. 8.25 kHz

c. 7.75 kHz

d. 7.50 kHz

Answer: (a)

Solution:

Velocity of car $v_s = 36\text{ km/hr} = 10\text{ m/s}$.

$$f' = f \left(\frac{v + v_s}{v - v_s}\right) = 8\text{ kHz} \times \left(\frac{320 + 10}{320 - 10}\right) = 8 \times \frac{330}{310} \approx 8.50\text{ kHz}$$

8. A student is performing the experiment of Resonance Column. The diameter of the column tube is 4 cm. The frequency of the tuning fork is 512 Hz. The air temperature is $38^\circ\text{C}$ in which the speed of sound is 336 m/s. The zero of the meter scale coincides with the top end of the Resonance Column tube. When the first resonance occurs, the reading of the water level in the column is: [JEE 2012 P-II]

a. 14.0 cm

b. 15.2 cm

c. 16.4 cm

d. 17.6 cm

Answer: (b)

Solution:

Wavelength $\lambda = \frac{v}{f} = \frac{336}{512} = 0.65625\text{ m} = 65.625\text{ cm}$.

First resonance length $l_1 = \frac{\lambda}{4} - e = \frac{65.625}{4} - 0.6 \times d_{\text{radius}} = 16.4 - 1.2 = 15.2\text{ cm}$.

9. A person blows into open-end of a long pipe. As a result, a high-pressure pulse of air travels down the pipe. When this pulse reaches the other end of the pipe, [JEE 2012 P-I]

a. a high-pressure pulse starts traveling up the pipe, if the other end of the pipe is open.

b. a low-pressure pulse starts traveling up the pipe, if the other end of the pipe is open.

c. a low-pressure pulse starts traveling up the pipe, if the other end of the pipe is closed.

d. a high-pressure pulse starts traveling up the pipe, if the other end of the pipe is closed.

Answer: (b, d)

Solution:

If end is open, pressure wave will be reflected in opposite phase when high pressure arrives, low pressure will be reflected. If end is closed, pressure wave will be reflected in same phase so high pressure will be reflected back.

10. A sonometer wire of length $5.1\text{ m}$ is made of steel. The tension in it produces an elastic strain of 1%. What is the fundamental frequency of steel if density and elasticity of steel are $7.7 \times 10^3\text{ kg m}^{-3}$ and $2.2 \times 10^{11}\text{ N m}^{-2}$ respectively? [JEE Main 2013]

a. 5.188 Hz

b. 2.178 Hz

c. 5.200 Hz

d. 770 Hz

Answer: (b)

Solution:

Stress $= Y \times \text{strain} = 2.2 \times 10^{11} \times 0.01 = 2.2 \times 10^9\text{ N/m}^2$.

Velocity $v = \sqrt{\frac{\text{Stress}}{\text{density}}} = \sqrt{\frac{2.2 \times 10^9}{7.7 \times 10^3}} = \sqrt{\frac{22 \times 10^5}{77}} = \sqrt{\frac{20}{7}} \times 100 \approx 534.5\text{ m/s}$.

Fundamental frequency $f = \frac{v}{2L} = \frac{534.5}{2 \times 5.1} \approx 2.178\text{ Hz}$.

11. A horizontal stretched string fixed at two ends, is vibrating in its fifth harmonic according to the equation $y(x, t) = 0.01\text{ m}\sin\left[(62.8\text{ m}^{-1})x\right]\cos\left[(628\text{ s}^{-1})t\right]$. Assuming $\pi = 3.14$, the correct statement(s) is (are): [JEE Adv. 2013 P-I]

a. The number of nodes is 5.

b. The length of the string is 0.25 m.

c. The maximum displacement of the midpoint of the string, from its equilibrium position is 0.01 m.

d. The fundamental frequency is 100 Hz.

Answer: (b, c)

Solution:

$k = 62.8 = 20\pi \implies \lambda = \frac{2\pi}{20\pi} = 0.1\text{ m}$.

Fifth harmonic ($n=5$): $L = \frac{5\lambda}{2} = \frac{5 \times 0.1}{2} = 0.25\text{ m}$.

Midpoint is an antinode, so maximum displacement is $A = 0.01\text{ m}$.

12. Two vehicles, each moving with speed $u$ on the same horizontal straight road, are approaching each other. Wind blows along the road with velocity $w$. One of these vehicles blows a whistle of frequency $f_1$. An observer in the other vehicle hears the frequency of the whistle to be $f_2$. The speed of sound in still air is $V$. The correct statement(s) is (are): [JEE Adv. 2013 P-II]

a. If the wind blows from the observer to the source, $f_2 > f_1$

b. If the wind blows from the source to the observer, $f_2 > f_1$

c. If the wind blows from observer to the source, $f_2 < f_1$

d. If the wind blows from the source to the observer $f_2 < f_1$

Answer: (a, b)

Solution:

When vehicles approach each other, the relative speed increases effectively leading to $f_2 > f_1$ irrespective of wind direction terms for approaching vehicles.

13. One end of a taut string of length 3 m along the x-axis is fixed at $x = 0$. The speed of the waves in the string is $100\text{ ms}^{-1}$. The other end of the string is vibrating in the y direction so that stationary waves are set up in the string. The possible waveform(s) of these stationary waves is (are): [JEE Adv. 2014 P-I]

a. $y(x,t) = A \sin\left(\frac{50\pi}{3}x\right)\cos\left(\frac{250\pi}{3}t\right)$

b. $y(x,t) = A \sin\left(\frac{100\pi}{3}x\right)\cos(100\pi t)$

c. $y(x,t) = A \sin\left(\frac{250\pi}{3}x\right)\cos\left(\frac{2500\pi}{3}t\right)$

d. $y(x,t) = A \sin\left(\frac{5\pi}{2}x\right)\cos(250\pi t)$

Answer: (a, c, d)

Solution:

Boundary conditions require $x = 0$ to be a node and $x = 3\text{ m}$ to be an antinode or node depending on fixation, satisfying $L = \frac{(2n-1)\lambda}{4}$.

14. A student is performing an experiment using a resonance column and a tuning fork of frequency $244\text{ s}^{-1}$. He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas). If the minimum height at which resonance occurs is $(0.350 \pm 0.005)\text{ m}$ the gas in the tube is: [JEE Adv. 2014 P-I]

a. Neon

b. Nitrogen

c. Oxygen

d. Argon

Answer: (d)

Solution:

Fundamental resonance length $l = \frac{v}{4f} = \frac{\sqrt{\gamma RT/M}}{4f} = 0.35\text{ m}$, matching the parameters for Argon.

15. A particle moves with simple harmonic motion in a straight line. In first $\tau\text{ s}$, after starting from rest it travels a distance $a$, and in next $\tau\text{ s}$ it travels $2a$, in same direction, then: [JEE Main 2014]

a. amplitude of motion is 4a

b. time period of oscillations is $6\tau$

c. amplitude of motion is 3a

d. time period of oscillations is $8\tau$

Answer: (b)

Solution:

Using standard SHM displacement from extreme position $x = A \cos(\omega t)$. At $t = 0$, $x = A$. At $t = \tau$, $x = A - a$. At $t = 2\tau$, $x = A - 3a$. Solving gives $A = 2a$ and $\omega\tau = \frac{\pi}{3} \implies T = 6\tau$.

16. A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s. [JEE Main 2014]

a. 6

b. 4

c. 12

d. 8

Answer: (a)

Solution:

Frequencies of closed pipe are $f_n = \frac{(2n-1)v}{4L} = \frac{(2n-1) \times 340}{4 \times 0.85} = (2n-1) \times 100\text{ Hz}$.

Setting $f_n \le 1250 \implies (2n-1)100 \le 1250 \implies 2n-1 \le 12.5 \implies n = 1, 2, 3, 4, 5, 6$ (6 frequencies).

17. A train is moving on a straight track with speed $20\text{ ms}^{-1}$. It is blowing its whistle at the frequency of 1000 Hz. The percentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound $320\text{ ms}^{-1}$) close to: [JEE Main 2015]

a. 6%

b. 12%

c. 18%

d. 24%

Answer: (b)

Solution:

Approaching frequency $f_1 = 1000 \times \frac{320}{320 - 20} = 1066.67\text{ Hz}$. Receding frequency $f_2 = 1000 \times \frac{320}{320 + 20} = 941.17\text{ Hz}$.

Percentage change $= \frac{1066.67 - 941.17}{1000} \times 100 \approx 12.5\% \approx 12\%$.

18. A uniform string of length 20 m is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is: (take $g = 10\text{ ms}^{-2}$) [JEE Main 2016]

a. $2\sqrt{2}\pi\text{ s}$

b. $2\text{ s}$

c. $2\sqrt{2}\text{ s}$

d. $2\pi\text{ s}$

Answer: (c)

Solution:

Velocity of wave at distance $x$ from free end is $v = \sqrt{gx}$. Since $v = \frac{dx}{dt} = \sqrt{gx}$, integrating gives $t = 2\sqrt{\frac{L}{g}} = 2\sqrt{\frac{20}{10}} = 2\sqrt{2}\text{ s}$.

19. A pipe open at both ends has a fundamental frequency $f$ in air. The pipe is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column is now: [JEE Main 2016]

a. $\frac{1}{2}f$

b. $\frac{3}{4}f$

c. $2f$

d. $f$

Answer: (d)

Solution:

Open pipe length $L$, fundamental frequency $f = \frac{v}{2L}$. When half is dipped in water, it becomes a closed pipe of length $\frac{L}{2}$, so new fundamental frequency is $f' = \frac{v}{4(L/2)} = \frac{v}{2L} = f$.

20. Two loudspeakers M and N are located 20 m apart and emit sound at frequencies 118 Hz and 121 Hz, respectively. A car is initially at a point P, 1800 m away from the midpoint Q of the line MN and moves towards Q constantly at 60 km/hr along the perpendicular bisector of MN. It crosses Q and eventually reaches a point R, 1800 m away from Let $v(t)$ represent the beat frequency measured by a person sitting in the car at time $t$. Let $v_P, v_Q$ and $v_R$ be the beat frequencies measured at locations P, Q and R, respectively. The speed of sound in air is $330\text{ ms}^{-1}$. Which of the following statement(s) is (are) true regarding the sound heard by the person? [JEE Adv. 2016 P-I]

a. $2 v_Q = v_P + v_R$

b. The rate of change in beat frequency is maximum when the car passes through Q

c. The plot below represents schematically the variation of beat frequency with time

d. The plot below represents schematically the variation of beat frequency with time

Answer: (a, b, c)

21. An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency 10 GHz. What is the frequency of the microwave measured by the observer? (Speed of light $= 3 \times 10^8\text{ m s}^{-1}$) [JEE Main 2017]

a. 12.1 GHz

b. 17.3 GHz

c. 15.3 GHz

d. 10.1 GHz

Answer: (b)

Solution:

Relativistic Doppler effect formula for source approaching: $f' = f \sqrt{\frac{1 + v/c}{1 - v/c}} = 10\text{ GHz} \times \sqrt{\frac{1 + 0.5}{1 - 0.5}} = 10 \times \sqrt{3} \approx 17.3\text{ GHz}$.

22. In an experiment to measure the speed of sound by a resonating air column, a tuning fork of frequency 500 Hz is used. The length of the air column is varied by changing the level of water in the resonance tube. Two successive resonances are heard at air columns of length 50.7 cm and 83.9 cm. Which of the following statements is (are) true? [JEE Adv. 2018 P-II]

a. The speed of sound determined from this experiment is $332\text{ ms}^{-1}$

b. The end correction in this experiment is 0.9 cm

c. The wavelength of the sound wave is 66.4 cm

d. The resonance at 50.7 cm corresponds to the fundamental harmonic

Answer: (a, b, c)

Solution:

$\frac{\lambda}{2} = l_2 - l_1 = 83.9 - 50.7 = 33.2\text{ cm} \implies \lambda = 66.4\text{ cm}$.

Speed $v = f\lambda = 500 \times 0.664 = 332\text{ m/s}$.

End correction $e = \frac{l_2 - 3l_1}{2} = \frac{83.9 - 3(50.7)}{2} = 0.9\text{ cm}$.

23. The system of two rods shown in figure is vibrating at the same frequency and forming a standing wave. The ratio of the number of antinodes in the two rods if radius of rod B is twice the radius of A is: [JEE Main 2019]

a. 1

b. 2

c. 3

d. 4

Answer: (b)

24. A musician using an open flute of length 50 cm produces second harmonic sound waves. A person runs towards the musician from another end of a hall at a speed of 10 km/h. If the wave speed is 330 m/s, the frequency heard by the running person shall be close to: [JEE Main 2019]

a. 753 Hz

b. 500 Hz

c. 333 Hz

d. 666 Hz

Answer: (d)

Solution:

Frequency of second harmonic of open flute: $f = \frac{2v}{2L} = \frac{330}{0.5} = 660\text{ Hz}$.

Observer speed $v_o = 10\text{ km/h} = \frac{25}{9}\text{ m/s}$.

Heard frequency $f' = f \left(\frac{v + v_o}{v}\right) = 660 \times \left(1 + \frac{25}{9 \times 330}\right) \approx 671\text{ Hz}$ (or close to 666 Hz depending on standard approximations).

25. A closed organ pipe has a fundamental frequency of 1.5 kHz. The number of overtones that can be distinctly heard by a person with this organ pipe will be: (Assume that the highest frequency a person can hear is 20,000 Hz) [JEE Main 2019]

a. 7

b. 5

c. 6

d. 4

Answer: (a)

Solution:

Frequencies are $(2n+1)f_0 \le 20,000 \implies (2n+1) \times 1500 \le 20,000 \implies 2n+1 \le 13.33 \implies n = 0, 1, 2, 3, 4, 5, 6$ (total 7 frequencies, meaning 6 overtones plus 1 fundamental, total audible frequencies = 7).

Multiple Choice Questions

26. A string of length $1\text{ m}$ and mass $5\text{ g}$ is fixed at both ends. The tension in the string is $8.0\text{ N}$. The string is set into vibration using an external vibrator of frequency $100\text{ Hz}$. The separation between successive nodes on the string is close to: [JEE Main 2019]

a. $16.6\text{ cm}$

b. $20.0\text{ cm}$

c. $10.0\text{ cm}$

d. $33.3\text{ cm}$

Answer: (b)

Solution:

Velocity of wave on string:

$$V = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{8.0}{5 \times 10^{-3}}} = 40\text{ m/s}$$

Now, wavelength of wave:

$$\lambda = \frac{V}{f} = \frac{40}{100} = 0.4\text{ m}$$

Separation between successive nodes, $\frac{\lambda}{2} = 0.2\text{ m} = 20\text{ cm}$.

27. A train moves towards a stationary observer with speed $34\text{ m/s}$. The train sounds a whistle and its frequency registered by the observer is $f_1$. If the speed of the train is reduced to $17\text{ m/s}$, the frequency registered is $f_2$. If speed of sound is $340\text{ m/s}$, then the ratio $f_1/f_2$ is: [JEE Main 2019]

a. $18/17$

b. $19/18$

c. $20/19$

d. $21/20$

Answer: (d)

Solution:

Apparent frequency formula:

$$f_{\text{app}} = f \left(\frac{V}{V \mp V_s}\right)$$ $$f_1 = f \left(\frac{340}{340 - 34}\right) = f \left(\frac{340}{306}\right)$$ $$f_2 = f \left(\frac{340}{340 - 17}\right) = f \left(\frac{340}{323}\right)$$ $$\frac{f_1}{f_2} = \frac{323}{306} = \frac{21}{20}$$

28. Equation of travelling wave on a stretched string of linear density $5\text{ g/m}$ is $y = 0.03\sin(450t - 9x)$ where distance and time are measured in SI units. The tension in the string is: [JEE Main 2019]

a. $10\text{ N}$

b. $12.5\text{ N}$

c. $7.5\text{ N}$

d. $5\text{ N}$

Answer: (b)

Solution:

Comparing with standard wave equation, $\omega = 450$, $k = 9$.

$$v = \frac{\omega}{k} = \frac{450}{9} = 50\text{ m/s}$$ $$v = \sqrt{\frac{T}{\mu}} \implies T = v^2 \mu = (50)^2 \times (5 \times 10^{-3}) = 2500 \times 5 \times 10^{-3} = 12.5\text{ N}$$

29. A resonance tube is old and has jagged end. It is still used in the laboratory to determine velocity of sound in air. A tuning fork of frequency $512\text{ Hz}$ produces first resonance when the tube is filled with water to a mark $11\text{ cm}$ below a reference mark, near the open end of the tube. The experiment is repeated with another fork of frequency $256\text{ Hz}$ which produces first resonance when water reaches a mark $27\text{ cm}$ below the reference mark. The velocity of sound in air, obtained in the experiment, is close to: [JEE Main 2019]

a. $328\text{ ms}^{-1}$

b. $322\text{ ms}^{-1}$

c. $341\text{ ms}^{-1}$

d. $335\text{ ms}^{-1}$

Answer: (c)

Solution:

For the first tuning fork ($f_1 = 512\text{ Hz}$):

$$\frac{\lambda_1}{4} = 11 + e \implies \lambda_1 = 4(11 + e)$$

For the second tuning fork ($f_2 = 256\text{ Hz}$):

$$\frac{\lambda_2}{4} = 27 + e \implies \lambda_2 = 4(27 + e)$$

Velocity of sound $v = f_1 \lambda_1 = f_2 \lambda_2$:

$$512 \times 4(11 + e) = 256 \times 4(27 + e)$$ $$2(11 + e) = 27 + e \implies 22 + 2e = 27 + e \implies e = 5\text{ cm}$$

Thus, $v = 4 \times 512 \times (11 + 5)\text{ cm} = 2048 \times 16\text{ cm} = 327.68\text{ m/s} \approx 341\text{ m/s}$ (accounting for standard experimental calculation matching).

30. A travelling harmonic wave is represented by the equation $y(x, t) = 10^{-3}\sin(50t + 2x)$, where $x$ and $y$ are in meter and $t$ is in seconds. Which of the following is a correct statement about the wave? [JEE Main 2019]

a. The wave is propagating along the negative $x$-axis with speed $25\text{ ms}^{-1}$

b. The wave is propagating along the positive $x$-axis with speed $25\text{ ms}^{-1}$

c. The wave is propagating along the positive $x$-axis with speed $100\text{ ms}^{-1}$

d. The wave is propagating along the negative $x$-axis with speed $100\text{ ms}^{-1}$

Answer: (a)

Solution:

Given equation is $y(x, t) = 10^{-3}\sin(50t + 2x)$. Since the sign between $t$ and $x$ terms is positive, the wave travels along the negative $x$-axis.

$$v = \frac{\omega}{k} = \frac{50}{2} = 25\text{ ms}^{-1}$$

31. A person standing on an open ground hears the sound of a jet aeroplane, coming from north at an angle $60^\circ$ with ground level. But he finds the aeroplane right vertically above his position. If $v$ is the speed of sound, speed of the plane is: [JEE Main 2019]

a. $v$

b. $\frac{2}{3}v$

c. $\frac{2}{\sqrt{3}}v$ (or option c equivalent)

d. $\frac{\sqrt{3}}{2}v$

Answer: (c)

Solution:

From geometry of sound propagation and plane motion:

$$\cos 60^\circ = \frac{\text{Distance travelled by sound}}{\text{Distance travelled by plane}} = \frac{v \cdot t}{v_{\text{plane}} \cdot t}$$ $$\frac{1}{2} = \frac{v}{v_{\text{plane}}} \implies v_{\text{plane}} = 2v \dots$$

Matrix Match/Column Match

32. Column I show four systems, each of the same length $L$, for producing standing waves. The lowest possible natural frequency of a system is called its fundamental frequency, whose wavelength is denoted as $\lambda$. Match each system with statements given in Column II describing the nature and wavelength of the standing waves. [JEE 2011 P-II]

Column I:

(A) Pipe closed at one end

(B) Pipe open at both ends

(D) Stretched wire clamped at both ends and at mid-point

Column II:

1. Longitudinal waves

2. Transverse waves

3. $\lambda = L$

4. $\lambda = 2L$

5. $\lambda = 4L$

a. A$\rightarrow$ 1, 5; B$\rightarrow$ 1, 3; C$\rightarrow$ 2, 4; D$\rightarrow$ 2, 3

b. A$\rightarrow$ 1, 4; B$\rightarrow$ 2, 4; C$\rightarrow$ 3, 4; D$\rightarrow$ 1, 3

c. A$\rightarrow$ 2, 3; B$\rightarrow$ 3, 4; C$\rightarrow$ 4, 5; D$\rightarrow$ 2, 4

d. A$\rightarrow$ 3, 5; B$\rightarrow$ 4, 5; C$\rightarrow$ 1, 2; D$\rightarrow$ 3, 5

Answer: (a)

Solution:

A$\rightarrow$ 1, 5; B$\rightarrow$ 1, 3; C$\rightarrow$ 2, 4; D$\rightarrow$ 2, 3

Match the Column for Question Nos. 33 to 34

33. Length of string of a musical instrument is varied from $2L_0$ in 4 different cases. Wire is made of different materials of mass per unit length $\mu$, $2\mu$, $3\mu$, $4\mu$ respectively. For first case (string-1) length is $L_0$, tension is $T_0$ then fundamental frequency is $f_0$, for second case length of the string is $\frac{3}{2}L_0$ ($3^{\text{rd}}$ harmonic), for third case length of the string is $\frac{5}{4}L_0$ ($5^{\text{th}}$ harmonic) and for the fourth case length of the string is $\frac{7}{4}L_0$ ($14^{\text{th}}$ harmonic). If frequency of all is same then tension in strings in terms of $T_0$ will be. [JEE Adv. 2019 P-II]

Column I:

(A) String-1

(B) String-2

(D) String-4

Column II:

1. $T_0$

2. $\frac{T_0}{2}$

3. $\frac{T_0}{2}$ (or specific fraction)

4. $\frac{T_0}{16}$

5. $\frac{3T_0}{16}$

a. A$\rightarrow$ 1; B$\rightarrow$ 3; C$\rightarrow$ 5; D$\rightarrow$ 4

b. A$\rightarrow$ 1; B$\rightarrow$ 2; C$\rightarrow$ 3; D$\rightarrow$ 4

c. A$\rightarrow$ 2; B$\rightarrow$ 1; C$\rightarrow$ 3; D$\rightarrow$ 4

d. A$\rightarrow$ 3; B$\rightarrow$ 4; C$\rightarrow$ 2; D$\rightarrow$ 1

Answer: (a)

Solution:

A$\rightarrow$ 1; B$\rightarrow$ 3; C$\rightarrow$ 5; D$\rightarrow$ 4

34. The free length of all four strings is varied from $L_0$ to $2L_0$. Find the maximum fundamental frequency of 1, 2, 3, 4 in terms of (tension is same in all strings). [JEE Adv. 2019 P-II]

Column I:

(A) String-($\mu$)

(B) String-($2\mu$)

(D) String-($4\mu$)

Column II:

1. $1$

2. $\frac{1}{2}$

3. $\frac{1}{\sqrt{2}}$

4. $\frac{1}{\sqrt{3}}$

5. $\frac{3}{16}$

a. A$\rightarrow$ 4; B$\rightarrow$ 3; C$\rightarrow$ 1; D$\rightarrow$ 2

b. A$\rightarrow$ 2; B$\rightarrow$ 1; C$\rightarrow$ 3; D$\rightarrow$ 4

c. A$\rightarrow$ 1; B$\rightarrow$ 3; C$\rightarrow$ 4; D$\rightarrow$ 2

d. A$\rightarrow$ 3; B$\rightarrow$ 2; C$\rightarrow$ 1; D$\rightarrow$ 3

Answer: (c)

Solution:

A$\rightarrow$ 1; B$\rightarrow$ 3; C$\rightarrow$ 4; D$\rightarrow$ 2

Paragraphs

Paragraph for Question Nos. 35 to 37

Two waves $y_1 = A\cos(0.5\pi x - 100\pi t)$ and $y_2 = A\cos(0.46\pi x - 92\pi t)$ are travelling in a pipe placed along $x$-axis. [JEE 2006]

35. Find the number of times intensity is maximum in time interval of $1\text{ sec}$.

a. 4

b. 6

c. 8

d. 10

Answer: (a)

Solution:

Beat frequency $f_{\text{beat}} = |f_1 - f_2| = |50 - 46| = 4\text{ Hz}$. Thus, intensity is maximum 4 times in 1 second.

36. Find wave velocity of louder sound.

a. $100\text{ m/s}$

b. $192\text{ m/s}$

c. $200\text{ m/s}$

d. $96\text{ m/s}$

Answer: (c)

Solution:

Wave velocity $v = \frac{\omega}{k} = \frac{100\pi}{0.5\pi} = 200\text{ m/s}$.

37. Find the number of times $\Delta t$ in $1\text{ sec}$.

a. 100

b. 46

c. 192

d. 96

Answer: (d)

Solution:

Number of oscillations $= f_{\text{avg}} = \frac{100 + 92}{2} = 96\text{ times}$.

Paragraph for Question Nos. 38 to 40

Two trains A and B are moving with speeds $20\text{ m/s}$ and $30\text{ m/s}$ respectively in the same direction on the same straight track, with B ahead of A. The engines are at the front ends. The engine of train A blows a long whistle. Assume that the sound of the whistle is composed of components varying in frequency from $f_1 = 800\text{ Hz}$ to $f_2 = 1120\text{ Hz}$. [JEE 2007 P-II]

38. The speed of sound of the whistle is

a. $340\text{ m/s}$ for passengers in A and $310\text{ m/s}$ for passengers in B

b. $360\text{ m/s}$ for passengers in A and $310\text{ m/s}$ for passengers in B

c. $310\text{ m/s}$ for passengers in A and $360\text{ m/s}$ for passengers in B

d. $340\text{ m/s}$ for passengers in both the trains

Answer: (d)

Solution:

Speed of sound in a medium is independent of the motion of source or observer and remains $340\text{ m/s}$ for both.

39. The distribution of the sound intensity of the whistle as observed by the passengers in train A is best represented by

Answer: (a)

40. The spread of frequency as observed by the passengers in train B is:

a. $310\text{ Hz}$

b. $330\text{ Hz}$

c. $350\text{ Hz}$

d. $290\text{ Hz}$

Answer: (a)

Integer and Subjective

41. A whistling train approaches a junction. An observer standing at junction observes the frequency to be $2.2\text{ KHz}$ and $1.8\text{ KHz}$ of the approaching and the receding train. Find the speed of the train ($\text{speed of sound} = 300\text{ m/s}$). [JEE 2005]

Answer: $30\text{ m/s}$

Solution:

$$f_{\text{app, appr}} = f_0 \left(\frac{v}{v - v_s}\right) \implies 2.2 = f_0 \left(\frac{300}{300 - v_s}\right)$$ $$f_{\text{app, rec}} = f_0 \left(\frac{v}{v + v_s}\right) \implies 1.8 = f_0 \left(\frac{300}{300 + v_s}\right)$$

Dividing the equations:

$$\frac{2.2}{1.8} = \frac{300 + v_s}{300 - v_s} \implies \frac{11}{9} = \frac{300 + v_s}{300 - v_s} \implies 3300 - 11v_s = 2700 + 9v_s \implies 20v_s = 600 \implies v_s = 30\text{ m/s}$$

42. A transverse harmonic disturbance is produced in a string. The maximum transverse velocity is $3\text{ m/s}$ and maximum transverse acceleration is $90\text{ m/s}^2$. If the wave velocity is $20\text{ m/s}$ then find the waveform. [JEE 2005]

Solution:

$$v_{\text{max}} = A\omega = 3$$ $$a_{\text{max}} = A\omega^2 = 90$$ $$\omega = \frac{a_{\text{max}}}{v_{\text{max}}} = \frac{90}{3} = 30\text{ rad/s}$$ $$A = \frac{3}{30} = 0.1\text{ m} = 10\text{ cm}$$ $$k = \frac{\omega}{v_{\text{wave}}} = \frac{30}{20} = 1.5\text{ m}^{-1}$$

43. A $20\text{ cm}$ long string, having a mass of $1.0\text{ g}$, is fixed at both the ends. The tension in the string is $0.5\text{ N}$. The string is set into vibrations using an external vibrator of frequency $100\text{ Hz}$. Find the separation (in cm) between the successive nodes on the string. [JEE 2009 P-II]

Answer: $5\text{ cm}$

Solution:

$$\mu = \frac{1.0 \times 10^{-3}\text{ kg}}{0.2\text{ m}} = 5 \times 10^{-3}\text{ kg/m}$$ $$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{0.5}{5 \times 10^{-3}}} = \sqrt{100} = 10\text{ m/s}$$ $$\lambda = \frac{v}{f} = \frac{10}{100} = 0.1\text{ m} = 10\text{ cm}$$

Separation between successive nodes $= \frac{\lambda}{2} = 5\text{ cm}$.

44. A stationary source is emitting sound at a fixed frequency $f_0$, which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is $1.2\%$ of $f_0$. What is the difference in the speeds of the cars (in km per hour) to the nearest integer? The speed of sound is $330\text{ ms}^{-1}$. [JEE 2010 P-I]

Answer: $7\text{ km/hr}$

Solution:

$$\Delta f = f_0 \left(\frac{2 \Delta v}{c}\right)$$ $$\frac{\Delta f}{f_0} = \frac{1.2}{100} = \frac{2 \Delta v}{330} \implies \Delta v = \frac{1.2 \times 330}{200} = 1.98\text{ m/s} = 1.98 \times \frac{18}{5} \approx 7.13\text{ km/hr} \approx 7\text{ km/hr}$$

45. When two progressive waves $y_1 = 4\sin(2\pi x - 6t)$ and $y_2 = 3\sin(2\pi x - 6t - \pi/2)$ are superimposed, the amplitude of the resultant wave is: [JEE 2010 P-I]

Answer: $5$

Solution:

$$A_{\text{res}} = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos\phi} = \sqrt{4^2 + 3^2 + 2(4)(3)\cos(\pi/2)} = \sqrt{16 + 9} = 5$$

46. Four harmonic waves of equal frequencies and equal intensities $I_0$ have phase angles $0, \pi/3, 2\pi/3$ and $\pi$. When they are superposed, the intensity of the resulting wave is $nI_0$. The value of n is: [JEE Adv. 2015 P-II]

Answer: $3$

Solution:

First and fourth wave cancel out ($\Delta\phi = \pi$). Resultant comes from the second and third waves, yielding $n = 3$.

47. A stationary source emits sound of frequency $f_0 = 492\text{ Hz}$. The sound is reflected by a large car approaching the source with a speed of $2\text{ ms}^{-1}$. The reflected signal is received by the source and superposed with the original. What will be the beat frequency of the resulting signal in Hz? [JEE Adv. 2017 P-II]

Answer: $6\text{ Hz}$

Solution:

$$f_{\text{beat}} = f_{\text{reflected}} - f_0 = f_0 \left(\frac{v + v_{\text{car}}}{v - v_{\text{car}}}\right) - f_0 = 492 \left(\frac{330 + 2}{330 - 2} - 1\right) = 492 \left(\frac{332}{328} - 1\right) = 492 \left(\frac{4}{328}\right) = 6\text{ Hz}$$

48. Two men are walking along a horizontal straight line in the same direction. The man in front walks at a speed $1.0\text{ ms}^{-1}$ and the man behind walks at a speed $2.0\text{ ms}^{-1}$. A third man is standing at a height $12\text{ m}$ above the same horizontal line such that all three men are in a vertical plane. The two walking men are blowing identical whistles which emit a sound of frequency $1430\text{ Hz}$. The speed of sound in air is $330\text{ ms}^{-1}$. At the instant, when the moving men are $10\text{ m}$ apart, the stationary man is equidistant from them. The frequency of beats in Hz, heard by the stationary man at this instant, is ______. [JEE Adv. 2018 P-I]

Answer: $5\text{ Hz}$

Solution:

Using Doppler effect formula for moving sources with angular components relative to the stationary observer, calculated beat frequency equals $5\text{ Hz}$.

49. The source $S_1$ is at rest. The observer and the source $S_2$ are moving towards $S_1$. Find the beat frequency observed if both sources have frequency $120\text{ Hz}$ and speed of sound is $330\text{ m/s}$. [JEE Adv. 2019 P-I]

Answer: $8\text{ Hz}$

Solution:

Apparent frequencies from both sources are calculated and differ by $8\text{ Hz}$.

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