Question

A tuning fork vibrating with a frequency of 512 𝐻𝑧 is kept close to the open end of a tube filled with water (figure). The water level in the tube is gradually lowered. When the water level is 17 𝑐𝑚 below the open end, maximum intensity of sound is heard. If the room temperature is ${20}^{\circ }C$,


$A$: Calculate, speed of sound in air at room temperature



$B$: Calculate, Speed of sound in air at $0^{\circ }C$



$C$: Calculate If the water in the tube is replaced with mercury, will there be any difference in your observations?

Answer 1

$A$:

Given,

A tuning fork vibrating with a frequency $=512Hz$

For first maximum intensity of sound is heard,

$L=\frac{\lambda }{4}\text{Lenght of the air column}$

$\Rightarrow \lambda =4L=4\times 17\times {10}^{-2}=0.68m$

Since, speed of sound, $v=v\lambda$

Substituting the values, we get
$v=512\times 0.68=348.16m/s$

Final Answer: $348.16m/s$


$B$:

Given,
Room temperature$={20}^{\circ }C$
speed of sound in air at ${20}^{\circ }C=348.16m/s$

Speed of sound in air at $0^{\circ }C$
$v\text{at}{20}^{\circ }C=v_1=348.16m/s$
$v\text{at}{0}^{\circ }C=v_2$

$\frac{v_1}{v_2}=\frac{{\sqrt{T}}_1}{{\sqrt{T}}_2}$
$\frac{348.16}{v_2}=\frac{\sqrt{273+20}}{\sqrt{273+0}}$
$v_2=348.16\times \frac{\sqrt{273}}{\sqrt{273+20}}$
$\frac{348.16}{1.03}=336m/s$

Final Answer: $336m/s$

$C$:

There will be no change in resonance length when water is replaced by mercury as it reflects the sound waves only. The intensity of sound may increase because mercury being denser than water will reflect sound waves more. So, intensity of sound heard will be longer, but reading does not change as medium in tube (air) and tuning fork are same.

Final Answer: No