Two tuning forks a and b of frequency 200 Hz and 400 Hz are vibrated simultaneously. speed of sound in air is 330 m/s. Then the ratio of time taken by the sound a and b to travel 660 m and 490 m respectively in air is ?

1 Answer
Jan 12, 2018

A) 2 seconds.

B) 1.49 seconds. (approx)

Explanation:

We know,

For a Sound wave,

c = nulambda.

Here, for Tuning Fork A,

Frequency nu = 200 Hz.

So, Time Period T = 1/200 seconds.

And from the equation c = nulambda, we get,

lambda = c/nu = (330)/200 m = 1.65 m

So, The Sound Wave emerging from Tuning Fork A covers,

1.65 m in rarr 1/200 seconds.

1 m in rarr 1/(200 * 1.65) seconds.

660 m in rarr 660/(2 * 165) seconds = 2 seconds.

Now, For Tuning Fork B,

Frequency nu = 400 Hz.

So, Time Period T = 1/400 seconds.

And from the equation c = nulambda, we get,

lambda = c/nu = 330/400 m = 0.825 m

So, The Sound Wave emerging from the Tuning Fork B covers,

0.825 m in rarr 1/400 seconds.

1 m in rarr 1/(400 * 0.825) seconds.

490 m in rarr 490/(4 * 82.5) seconds approx 1.49 seconds.