Question

A tuning fork produces four beats per second
with 49cm and 50cm lengths of a stretched
wire of a sonometer. The frequency of the fork
is

1) 196 Hz
2) 296 Hz
3) 396 Hz
4) 693 Hz​

Answer 1

Answer:

3) 396 Hz

Step-by-step explanation:

Let's say the frequency of the fork is f

and the speed of wave on the string is v

then according to question it gives 4 beats/s with the stretched wires of length 48 cm and 50 cm.

therefore,

v/2*48-f =4  . . . . step(1)

and

f-v/2*50=4  . . . . step(2)

on solving both the equations we get

f=396 HZ

Answer 2

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3) 396 Hz

Solution:

$n = \frac{1}{2l} \sqrt{ \frac{t}{m} }$

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

$\frac{ n_{1} }{ n_{2} } = \frac{50}{49}$

In Both cases wire producing 4 beats with

fork. This means the time string is 4 Hz higher

and in other case 4 Hz lower.

$n_{1} - n_{2} = 8$

$\frac{ n_{1} }{ n_{2} } = \frac{50 \times 8}{49 \times 8} = \frac{400}{392}$

$\frac{ n_{1} }{ n_{2} } =396 \: hz$