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) 22 seconds.

B) 1.491.49 seconds. (approx)

Explanation:

We know,

For a Sound wave,

c = nulambdac=νλ.

Here, for Tuning Fork A,

Frequency nuν = 200200 Hz.

So, Time Period TT = 1/2001200 seconds.

And from the equation c = nulambdac=νλ, we get,

lambda = c/nu = (330)/200λ=cν=330200 m = 1.651.65 m

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

1.651.65 m in rarr 1/2001200 seconds.

11 m in rarr 1/(200 * 1.65)12001.65 seconds.

660660 m in rarr 660/(2 * 165)6602165 seconds = 22 seconds.

Now, For Tuning Fork B,

Frequency nuν = 400400 Hz.

So, Time Period TT = 1/4001400 seconds.

And from the equation c = nulambdac=νλ, we get,

lambda = c/nu = 330/400λ=cν=330400 m = 0.8250.825 m

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

0.8250.825 m in rarr 1/4001400 seconds.

11 m in rarr 1/(400 * 0.825)14000.825 seconds.

490490 m in rarr 490/(4 * 82.5)490482.5 seconds approx 1.491.49 seconds.