A tuning fork vibrating with a frequency of $512$ $\mathrm{Hz}$ is kept close to the open end of a tube filled with water. The water level in the tube is gradually lowered. When the water level is $17$ $\mathrm{cm}$ below the open end, maximum intensity of sound is heard. If the room temperature is $20^{°}$ $\mathrm{C}$, calculate :
$(a)$ speed of sound in air at room temperature
$(b)$ speed of sound in air at $0^{°}$ $\mathrm{C}$.
$(c)$ if the water in the tube is replaced with mercury, will there be any difference in your observations ?
A tuning fork vibrating with a frequency of $512$ $\mathrm{Hz}$ is kept close to the open end of a tube filled with water. The water level in the tube is gradually lowered. When the water level is $17$ $\mathrm{cm}$ below the open end, maximum intensity of sound is heard. If the room temperature is $20^{°}$ $\mathrm{C}$, calculate :
$(a)$ speed of sound in air at room temperature
$(b)$ speed of sound in air at $0^{°}$ $\mathrm{C}$.
$(c)$ if the water in the tube is replaced with mercury, will there be any difference in your observations ?
$(a)$ For maximum intensity of sound wave at open end of closed pipe, we have (in the first
mode) $L=\frac{\lambda}{4}$
$\therefore \lambda=4 \mathrm{~L}=4 \times 0.17=0.68 \mathrm{~m}$
Now, $v=f \lambda=(512)(0.68)=348.16 \mathrm{~m} / \mathrm{s}$
$(b)$ Speed of sound in air is $v \propto \sqrt{\mathrm{T}}$
$\therefore \frac{v_{1}}{v_{2}}=\sqrt{\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}} \quad$ (Where $v_{1}=$ speed of
sound in air at $0^{\circ} \mathrm{C}$ temp.)
$\therefore \frac{v_{1}}{348.16}=\sqrt{\frac{0+273}{20+273}}$
$\therefore v_{1}=348.16 \times \sqrt{\frac{273}{293}} \approx 336 \frac{\mathrm{m}}{\mathrm{s}}$
$(c)$ Mercury is $13.6$ times denser than water. Hence its surface has reflectively much more than that of water. Hence, when water is replaced by mercury, we get almost total reflection of sound waves from its surface. Hence, we get greater intensity of reflected sound as com pared to water surface.
But certainly wavelength and speed of reflected sound wave will remain same.
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