Question

Two tuning forks 'P' and 'Q' when sounded together produce 10 beatsper second. On loading fork 'P' with a little wax, when soundedtogether, forks 'P' and 'Q' produce 4 beats per second. If the frequencyof fork 'Q' is 620 Hz., determine the frequency of fork P.​

Answer 1

Given :

Beats per second without covering wax = 10

Beats per second with covering wac on fork P n = 4

Frequency of fork Q N = 620 Hz

To find :

Frequency of fork P

Solution :

When two waves of approximately equal frequencies start vibrating and travelling in a medium in same direction then these waves superimpose .As a result a resultant sound appears whose amplitude increases and decreases regularly.

1.The phenomena of beats is used in tuning two vibrating bodies in unison. Like, a sonometer wire, it can be tuned in unison with a tuning fork by observing the beats. When an excited tuning fork is kept on the sonometer and if the sonometer wire is also excited,we can listen beats when the frequencies are approx equal. If the length of the wire is adjusted carefully to decrease the number of beats to zero, then the two are called in unison.

2. Beats can be used to figure out the frequency of a tuning fork. A standard tuning fork of frequency N is excited with the experimental fork. If n is the number of beats per second .Then frequency of experimental fork is N+n. The experimental fork is then loaded with a little wax, due to which frequency decreases. Now the observations are repeated. If the number of beats increases, then the frequency of the experimental fork is N-n, and if the number of beats decreases its frequency is N + n.

In this case the number of beats decrease so,

frequency = N + n

When wax is applied on fork P, then its frequency will decrease so number of beats will also decrease.

So

Frequency of fork P = N + n

as n = 4 and N = 620 Hz

so

Frequency of fork P = 620 + 4 = 624 Hz

Answer 2

The frequency of tuning fork P will be 630 Hz

Explanation:

Given,

Frequency of Q = 620 Hz

Since P and Q produce 10 beats per second

Therefore, the frequency of P

$=620\pm10$ Hz

$=630 \text{Hz }\text{or }610\text{Hz}$

We know that when a tuning fork is loaded is wax, its frequency decreases

Therefore, the frequency of P must have decreased

Now, if we take frequency of P as 610 Hz, then after loading P with wax, its frequency will further decrease

As a result, the beat frequency will increase.

But given that beat frequency is 4 beats/second

Hence, beat frequency is decreasing

Therefore, the frequency of P must be 630 Hz

Hope this answer is helpful.

Know More:

Q: Two tuning forks have frequencies 412 Hz and 409 Hz respectively.Calculate the period of the beat.

Click Here: https://brainly.in/question/8355419

Q: Two tuning forks A and B when sounded together give 4 beats per second. The fork A is in resonance with a closed column of air of length 15 cm, while the second is in resonance with an open column of length 30.5 cm.Calculate their frequencies.

Click Here: https://brainly.in/question/191273