Question

A wire of length 'L' and linear density 'm' is stretched between two rigid supports with tension 'T'. It is observed that wire resonates in the $p^{th}$ harmonic at a frequency of 320 Hz and resonates again at next higher frequency of 400 Hz. The value of 'p' is

• A. 2
• B. 8
• C. 4
• D. 10

Answer 1

Answer: (C)

4

Answer 2

The frequency of the $n^{th}$ harmonic of a stretched wire is given by

$$f_n = \frac{n}{2L}\sqrt{\frac{T}{m}},$$

where $n$ is the harmonic number.

Given:

$$f_p = 320\,\text{Hz}, \quad f_{p+1} = 400\,\text{Hz}.$$

Taking the ratio:

$$\frac{f_{p+1}}{f_p} = \frac{p+1}{p} = \frac{400}{320} = \frac{5}{4}.$$

Thus,

$$\frac{p+1}{p} = \frac{5}{4} \quad \Rightarrow \quad 4(p+1)=5p \quad \Rightarrow \quad 4p+4=5p \quad \Rightarrow \quad p=4.$$