Question

2 tuning forks A and B of frequencies 200 Hz and 400 Hz are vibrated simultaneously. Then the ratio of time taken by the sound produced by A and B to travel 660 m and 490 m respectively in air is ________(velocity of the sound in air = 330 m/s) A. 1:2 B. 1:3 C. 2:3 D 1:1

Answer 1

Let's start with Fork A:

The frequency of vibration is 200 Hz and the speed of sound in air is 330 m/s. In order to get the time taken by the sound to travel a certain distance, we first need to get the wavelength of the sound wave.

$v=f*\lambda$; where v is the velocity of sound in air, f is the frequency of the wave and $\lambda$ is the wavelength.

Therefore, $\lambda=\frac{330}{200}=1.65\:m$

Which means that the wave travels 1.65 m in $\frac{1}{200}$sec ($time=1/frequency$).

Therefore, wave A will travel the 660 m in $\frac{660m}{200/sec*1.65m}\\=2\:seconds$.

Similarly, for Fork B:

The frequency of vibration is 400 Hz and the speed of sound in air is 330 m/s.

$\lambda=\frac{330}{400}=0.85\:m$

Which means that the wave travels 0.85 m in $\frac{1}{400}$sec ($time=1/frequency$).

Therefore, wave B will travel the 490 m in $\frac{490m}{400/sec*0.85m}\\=1.5\:seconds$.

Therefore, the answer is 2 : 1.5 (*2)

Answer = 4 : 3

Answer 2

Explanation:

Let's start with Fork A:

The frequency of vibration is 200 Hz and the speed of sound in air is 330 m/s. In order to get the time taken by the sound to travel a certain distance, we first need to get the wavelength of the sound wave.

v=f*\lambdav=f∗λ ; where v is the velocity of sound in air, f is the frequency of the wave and \lambdaλ is the wavelength.

Therefore, \lambda=\frac{330}{200}=1.65\:mλ=

200

330

=1.65m

Which means that the wave travels 1.65 m in \frac{1}{200}

200

1

sec (time=1/frequencytime=1/frequency ).

Therefore, wave A will travel the 660 m in \begin{lgathered}\frac{660m}{200/sec*1.65m}\\=2\:seconds\end{lgathered}

200/sec∗1.65m

660m

=2seconds

.

Similarly, for Fork B:

The frequency of vibration is 400 Hz and the speed of sound in air is 330 m/s.

\lambda=\frac{330}{400}=0.85\:mλ=

400

330

=0.85m

Which means that the wave travels 0.85 m in \frac{1}{400}

400

1

sec (time=1/frequencytime=1/frequency ).

Therefore, wave B will travel the 490 m in \begin{lgathered}\frac{490m}{400/sec*0.85m}\\=1.5\:seconds\end{lgathered}

400/sec∗0.85m

490m

=1.5seconds

.

Therefore, the answer is 2 : 1.5 (*2)

Answer = 4 : 3