Question

A tuning fork and air whose temperature is ${51}^{o}C$ produce $4$ beats in one second when sounded together. When the temperature of air column decreases. When the temperature remains ${16}^{o}C$ only one beats per second is produced. the frequency of the tuning is

• A. $100Hz$
• B. $75Hz$
• C. $150Hz$
• D. $50Hz$

Answer 1

The correct option is D $50Hz$

Frequency of sound wave is given as :

$\nu =\frac{v}{\lambda }$

where $v$ is the speed of sound wave. If the temperature increases, the energy of molecules increases and hence the molecules vibrate ore faster so the speed of sound wave increases and vice versa.

Hence, on decreasing the temperature the speed of sound wave also decreases.

Let $\nu$ be the frequency of tuning fork. So, at ${51}^{0}C$ the frequency of air column is $(n+4)$ and at ${16}^{0}C$ the frequency is $(n+1)$.

So,

${\nu }_{51}=(n+4)\lambda ..........\left(1\right)$

${\nu }_{16}=(n+1)\lambda ...........\left(2\right)$

$dividing\left(1\right)and\left(2\right)$

$\frac{n_{51}}{n_{16}}=\frac{(n+4)}{(n+1)}$

$\because v\propto \sqrt{T}$

$so,\sqrt{\frac{273+51}{273+16}}=\frac{n+4}{n+1}$

$\sqrt{\frac{324}{289}}=\frac{n+4}{n+1}$

$\frac{18}{17}=\frac{n+4}{n+1}$

$18n+18=17n+68$

$n=50Hz$