Question

A tuning fork of frequency 500 Hz is sounded on a resonance tube. The first & second resonances are obtained at 17 cm and 52 cm. The velocity of sound in m/s is

Answer 1

Answer:

350 m/s

Step by step explanation:

The first resonance occurs at $\lambda/4$, while the second resonance occurs at $3\lambda/4$.

Therefore,

$17+e=\lambda/4$ and $52+e=3\lambda/4$.

If we subtract the two equations from each other:

$52-17=\lambda/2\\\lambda=70\:cm=0.7\:m$.

Hence,

$v=f*\lambda\\v=500*0.7=350\:m/s$

Answer 2

Answer:

The velocity of sound is 350 m/s.

Explanation:

Given that,

Frequency f = 500 Hz

First resonance = 17 cm

Second resonance = 52 cm

First resonant length

$\dfrac{\lambda}{4}=L_{1}+e$

Here, $\lambda$= wave length

L₁=length of first resonant

$\dfrac{\lambda}{4}=17+e$....(I)

Second resonant length

$\dfrac{3\lambda}{4}=L_{2}+e$

Here, L₂=length of first resonant

$\dfrac{3\lambda}{4}=52+e$....(II)

Subtract equation (I) from equation (II)

$\dfrac{3\lambda}{4}-\dfrac{\lambda}{4}=52+e-17-e$

$\dfrac{\lambda}{2}=35$

$\lambda = 70 cm$

The velocity of sound will be

$v= \lambda \times f$

$v= 0.7\times 500$

$v= 350\ m/s$

Hence, The velocity of sound is 350 m/s.