Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# A girl stops singing a pure note, she is surprised to hear an echo of higher frequency, that is, a higher musical pitch. Then: A) there could be some warm air between the girl and the reflecting surface.B) there could be two identical fixed reflecting surfaces, one half a wavelength of the sound wave away from the other.C) the girl could be moving towards a fixed reflector or vice versa.D) there could be less dense medium between the girl and the reflecting surface.

Last updated date: 20th Jun 2024
Total views: 55.5k
Views today: 1.55k
Verified
55.5k+ views
Hint: We know that the frequency of sound is directly related to velocity. if velocity increases the frequency increases.In the case where there is relative motion between the source and the observer the doppler effect should be used to find the frequency of the reflected sound. Using this we can check which of the options is correct.

Complete step by step solution:
It is given that a girl singing a pure note hears an echo with a higher frequency. We need to find the reason behind this increase in frequency.
Let us analyze the options one by one.
In the first option it is given that the increase in frequency can be due to the presence of warm air between the girl and reflector. When the air is warm it means the sound wave can travel with a higher velocity. when temperature increases the molecules vibrate faster so the sound travels faster.
The relation between velocity and frequency of sound is given as
$v = f\lambda$
Where v is the velocity, f is the frequency and $\lambda$ is the wavelength.
So, higher velocity means higher frequency. Which means the original frequency will be increased and the reflected frequency will also be increased. But they both will be increased to the same amount. So, the reflected frequency will not be heard as a higher one. It will be the same as that of the increased original frequency.

Thus, option A is wrong.

In the second option it is given that there could be two identical fixed reflecting surfaces, one half a wavelength of the sound wave away from the other. The frequency does not depend upon the distance it only depends upon the velocity.

Hence this option is wrong.

In the option C it is given that there is a relative motion between the reflector and the girl.
we know that according to Doppler effect the received frequency is given as
$f' = f\left( {\dfrac{{v + {v_0}}}{{v - {v_s}}}} \right)$
Where $f$ is the initial frequency $v$ is the velocity of sound ${v_0}$ is the velocity if the observer and ${v_s}$ is the velocity of the source.
So, when the girl moves towards the reflector or vice versa we can see that the value of reflected frequency increases.

So, option C is the correct answer.

Option D is also not correct because velocity of sound decreases in less dense medium. And the decrease will be equal for original and reflected sound.

So, the only correct answer is option C.

Note: Remember that sound is a longitudinal wave. It requires a medium to travel. In the case of light we know that the light slows down when it moves through a denser medium but in the case of sound the velocity increases when it moves through a denser medium. And with increase in velocity frequency also increases. The velocity of sound also increases with temperature since the molecules vibrate faster when it is hot.