Q125 / 150 · MHT CET 2026 (18 April Shift 2) · Physics

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 in two successive modes. The value of 'P' is

  • A
    $2$
  • Correct answer
    B
    $4$
  • C
    $8$
  • D
    $10$

Solution

The frequency of the $n^{th}$ harmonic of a stretched wire fixed at both ends is given by $f_n = n f_1$, where $f_1$ is the fundamental frequency.

For the $P^{th}$ harmonic, the frequency is given as:

$f_P = P f_1 = 320$

The next higher resonant frequency corresponds to the $(P+1)^{th}$ harmonic:

$f_{P+1} = (P+1) f_1 = 400$

Dividing the two equations, we get:

$\dfrac{P+1}{P} = \dfrac{400}{320}$

$\dfrac{P+1}{P} = \dfrac{5}{4}$

$5P = 4P + 4$

$P = 4$

Answer: $4$

Hence, the correct answer is option: 'B'.

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