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This is second of our many quizzes on Sound. Click on this to access more such competitive quizzes on SHM Waves and Optics.

Sound Quiz 2

physics

Aditya is very curious about gases so he decided to calculate the speed of sound in a certain diatomic gas with $\gamma =1.4$. The density of gas at NTP is $\rho =1.36Kg/{{m}^{3}}$. The speed of sound in this gas will be-$[\text{pressure at NTP}\,1\times {{10}^{5}}N/{{m}^{2}}]$. 

$\gamma $ is an adiabatic constant and depends on the nature of the gas. 

 Speed of sound in a gas is given by, $ \\  v=\sqrt{\dfrac{\gamma P}{\rho }}\\  \Rightarrow v=\sqrt{\dfrac{1.4\times 1\times {{10}^{5}}}{1.36}}\\  \therefore v =320.8\simeq 320m/s$ 

Ramesh and Suresh are very close friends, one day they decided to play their music on the speaker at home and decided to walk in the opposite direction. Ramesh starts walking from speaker to north direction and Suresh starts walking towards the south at the same speed at the same time. The wind is flowing towards north direction, which of the following statements is correct-

If observers are moving with the same speed away from the stationary source, then the frequency is equal. 

 Ramesh and Suresh are moving observers and the speaker at home is a stationary source, both are moving at the same speed and opposite directions of the source so, the frequency received by them will also be equal. 

Ravin is travelling in a car with a speed of 5 m/s towards a wall, during his journey, he opened the window of the car and he observed the wind direction which is also flowing in his direction of motion. Now, he is curious to know about echoes from a certain wall so when the car is 100 m away from the wall, he sounds a whistle from which an echo is heard by him after travelling some distance X at point P. The distance X will nearly be equal to-$[\text{speed of sound is 345m/s}]$<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/fde035ea-b53d-4342-8b8c-2cfdf6faac5b6407626661756909067.png"/> 

The time taken by the car to reach point $P$ is equal to the time taken by sound to travel 100 m and return to point P as echo is heard at that point. 

The time taken by the sound in moving 100 m from whistle point and coming back, by travelling 100-X m will be given by, $ \\  {{t}_{1}}=\dfrac{100}{V}+\dfrac{100-X}{V}\\ $ Time taken by car to move from whistle point to distance X will be $ \\  {{t}_{2}}=\dfrac{X}{u}\\ $ As the car and echo reaches point P at same time value ${{t}_{1}}\,\text{and}\,{{t}_{2}}$ will be equal so,$ \\ \dfrac{X}{u}=\dfrac{100}{V}+\dfrac{100-X}{V}\\  \Rightarrow \dfrac{X}{5}=\dfrac{100}{345}+\dfrac{100-X}{345}\\  \Rightarrow 70X=200\\  \therefore X\simeq 3$  

A 3D sound speaker is placed at a point due to which it emits an average power output of 120W. Ram is standing at a distance of 6m from the source speaker. The intensity received at Ram’s location due to speaker will be- 

Intensity is distributed in the form of a spherical surface on which at constant radius the intensity is the same on the surface spherical distance of $r$. 

Power output at a distance ‘r’ is given by,$ \\  P=4\pi {{r}^{2}}I\\ $ Now, putting the values from problem we have, $ \\  120=4\pi {{6}^{2}}I\\  \Rightarrow I=\dfrac{120}{4\pi \times 36}\\  \therefore I=0.265W/{{m}^{2}}$ 

 Amit loves experimenting, he decided to place his Bluetooth speaker at a frequency of 500Hz on a rotating block rotating with angular velocity $15rad/s$ and the radius of circular motion is 2m as shown in the figure. Now he is standing at some distance as shown in fig., calculate the maximum frequency received by Amit in one complete circular motion of speaker-$[\text{speed of sound is 350m/s}]$<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/bfffc5e3-85ed-4fdb-a7d3-5b4968d353d05862104913872757658.png"/> 

 In other cases, the direction of the velocity of the speaker will be at some angle from the observer so the value will be decreased. So, it will be maximum when the observer or source or both will be moving in a straight line towards each other. 

The maximum frequency received by Amit at a given position will be as shown in figure. As the speaker is directly approaching towards Amit, and the new frequency will be given by $ \\  f'=\left( \dfrac{V}{V-u} \right)f\\ $ Tangential velocity of speaker is, $ \\  u=\omega r=15\times 2=30\,m/s\\ $ So, maximum frequency will be $ \\  f'=\left( \dfrac{350}{350-30} \right)500\\ \therefore f' =547Hz$