Question

A sound source of frequency 170 Hz is placed near a wall. A man walking from the source normally towards the wall finds that there is a periodic rise and fall of sound intensity. If speed of sound in air is 340 m/s, the distance in meters separating the two adjacent position of minimum intensity is:
A. \[\dfrac{1}{2}\]
B. 1
C. \[\dfrac{3}{2}\]
D. 2

Answer 1

Hint: In this question we have been asked to calculate the position of minimum intensity of a sound wave with frequency of 170 Hz. It is given that a man walking near the sound source hears a periodic rise and fall of sound intensity. We know that distance separating the position of minimum frequency is given as half the wavelength of the sound wave. Therefore, to calculate the position of minimum sound intensity we shall calculate the wavelength of the given sound wave.

Formula used:
\[v=n\lambda \]
Where,
v is the velocity of sound
n is the frequency of the wave
and \[\lambda \] is the wavelength,

Complete step-by-step answer:
It is given that the frequency of the sound wave is 170 Hz. The speed of sound is given as 340 m/s.
Now, from the formula, we know,
\[v=n\lambda \]
Solving for \[\lambda \]
We get,
\[\lambda =\dfrac{v}{n}\]
After substituting given values
We get,
\[\lambda =\dfrac{340}{170}\]
Therefore,
\[\lambda =2m\]……………. (1)
Now, we also know that, distance separating the two position of minimum intensity (D) is given by,
\[D=\dfrac{\lambda }{2}\]
Substituting from (1)
We get,
\[D=1m\]

So, the correct answer is “Option B”.

Note: The power carried by sound wave per unit area in a direction that is perpendicular to that area is known as sound intensity. Sound intensity is also known as acoustic intensity. Sound intensity is given as the product of sound pressure and the velocity of the particle or the sound wave. The SI unit of sound intensity is Watt per square metre.