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 ?

Question

5721 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-12T08:06:22

A) #2# seconds.

B) #1.49# seconds. (approx)

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.