Question

A sonometer wire 100 cm long produces a resonance with a tuning fork. When its length is decreased by 10cm, 8 beats per second are heard. Find the frequency of tuning fork. (Ans : 72 Hz.)

Answer 1

we know, number of beats per second = difference in frequencies.

$8=n_2-n_1$.....(1)


given, beats per second = 8
inital length of wire , L = 100cm
final length of wire , l = 100cm - 10cm = 90cm
speed of sound , v = 340 m/s

we know, frequency is inversely proportional to length of wire.

so, $\frac{n_1}{n_2}=\frac{l_2}{l_1}$

or, $\frac{n_1}{n_2}=\frac{90}{100}=\frac{9}{10}$

$n_2=\frac{10}{9}n_1$......(2)

from equations (1) and (2),

$\frac{10}{9}n_1-n_1=8$

or, $\frac{n_1}{9}=8$

or, $n_1=8\times9$ = 72Hz

hence, frequency of tunning fork is 72Hz

Answer 2

Answer:

The frequency of tuning fork is 72 Hz.

Explanation:

Given That,

Initial length of wire $l_{1} = 100\ cm$

Final length of wire $l_{2} = 100-10=90\ cm$

We know that,

The number of beat per second $N=n₂-n₁$

$8=n_{2}-n_{1}$....(I)

Using the formula

$n= \dfrac{1}{2l}\sqrt{\dfrac{T}{m}}$

Since,

$n_{1}\propto\dfrac{1}{l_{1}}$

$n_{2}\propto\dfrac{1}{l_{2}}$

Therefore,

$\dfrac{n_{1}}{n_{2}}=\dfrac{l_{2}}{l_{1}}$

$\dfrac{n_{1}}{n_{2}}=\dfrac{90}{100}$

$n_{2}=\dfrac{100}{90}n_{1}$....(II)

Put the value of n₁ in equation (I)

$8=\dfrac{100}{90}n_{1}-n_{1}$

$n_{1}= 72\ Hz$

Hence, The frequency of tuning fork is 72 Hz.