Question

A tuning fork of frequency 100Hz when sounded together with another tuning fork of unknown
frequency produces 2 beats per second. On loading the tuning fork whose frequency is not known and sounded together with a tuning fork of frequency 100Hz produces one beat, then the frequency
of the other tuning fork is​

Answer 1

this is your answer SIR

hope it helps mark me as brainliest thanks

and give me a lot of thanks

Answer 2

Question:

A tuning fork of frequency 100Hz when sounded together with another tuning fork of unknown

frequency produces 2 beats per second. On loading the tuning fork whose frequency is not known and sounded together with a tuning fork of frequency 100Hz produces one beat, then the frequency of the other tuning fork is.

Given:

Frequency of known tuning fork = 100Hz

In first case:

• beats/second = 2

Let the frequency of another unknown tuning fork be $\nu_1$

$\sf No.~of ~beats = | \nu_1 - 100 | \\\\ \sf 2 = | \nu_1 - 100| \\\\ \sf \nu_1 = 98~or~102$

Now, it's then loaded with wax,

Let the frequency of the other tuning fork after loading is $\nu_2$

$\sf No.~of ~beats = | \nu_2 - 100| \\\\ \sf 1 = | \nu_2 - 100| \\\\ \sf \nu_2= 99~or~101$

So, here we have two possibilities for $\nu_1$ and $\nu_2$

Note: The purpose of loading is to decrease the frequency of the tuning fork.

Since, we know that, the frequency of tuning fork reduces after loading, $\nu_2$ should be less than $\nu_1$

We have $\nu_1 = 98~or~102$

If we take $\nu_1 = 98$, we will end up having $\nu_2$ greater than $\nu_1$ i.e 99 or 101.

So, to comply with the rule, we have to take $\nu_1 = 102$, so that we get value of $\nu_2$ lesser than $\nu_1$ i.e 101.