Question

Two tuning forks P and Q are vibrated together. The number of beats produced are represented by the straight line OA in the following graph. After loading Q with wax again these are vibrated together and the beats produced are represented by the line OB. If the frequency of P is 341Hz, the frequency of Q will be



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Answer 1

Answer:

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Answer 2

The frequency of Q will be 344 Hz

Explanation:

The figure is attached.

Given,

Frequency of P = 341 Hz

According to the graph for 1 b/s, the beat frequency is 3 Hz

Therefore, the frequency of Q

$=341\pm3$ Hz

$=338 \text{Hz }\text{or }344\text{Hz}$

From the graph it is clear that for 1 b/s, the beat frequency is reducing

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

Therefore, the frequency of Q must have decreased

Now, if we take frequency of Q as 338 Hz, then after loading Q with wax, its frequency will further decrease

As a result, the beat frequency will increase.

But according to the graph, beat frequency is decreasing

Therefore, the frequency of Q must be 344 Hz

Hope this answer is helpful.

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