A fork and a string of length 100 cm give 4 beats/ s . The string is made shorter, without any change of tension and linear mass density until it is in unison with the fork. It s new length is 99 cm .

Question

A fork and a string of length 100 cm give 4 beats/ s . The string is made shorter, without any change of tension and linear mass density until it is in unison with the fork. It s new length is 99 cm . The frequency of the fork is 80 P Hz . Find the value of P (Assume string to be vibrating in fundamental mode only)

Answer:

5

Solution:

$f = \frac{1}{2 \times 99} \frac{F}{μ}$ and $f - 4 = \frac{1}{2 \times 100} \frac{F}{μ} \Rightarrow \frac{f}{f - 4} = \frac{100}{99} f = 400 Hz$

TestHub · Accepted Answer

f = 2 × 99 1 ​ μ F ​ ​ and f − 4 = 2 × 100 1 ​ μ F ​ ​ ⇒ f − 4 f ​ = 99 100 ​ f = 400 Hz

Physics - Sound Wave Question with Solution | TestHub

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