wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A source of sound S and detector D are placed at some distance from one another. a big cardboard is placed near hte detector and perpendicular to the line SD as shown in figure. It is gradually moved away and it is found that the intensity changes from a maximum to a minimum as the board is moved through a distance of 20 cm. Find the frequency of the sound emitted. Velocity of sound in air is 336 m s−1.

Open in App
Solution

Given:
Velocity of sound in air v = 336 ms−1
Distance between maximum and minimum intensity: λ4 = 20 cm
Frequency of sound f = ?

We have:
λ4=20λ=20×4=80 cm=80×10-2 m

As we know, v=fλ.
f=vλ
f=33680×10-2=420 Hz

Therefore, the frequency of the sound emitted from the source is 420 Hz.

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Amplitude
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon