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MHT CET 2024 (16 May, Shift 1) · Stationary Waves & Vibrating Strings

Stationary Waves & Vibrating Strings question from MHT CET 2024 (16 May, Shift 1)

Q48MHT CET 2024 · 16 MAY · SHIFT 1Medium

Stationary Waves & Vibrating Strings

A string is under tension of 180 N and mass per unit length 2×10−3 kg/m2 \times 10^{-3}\text{ kg/m}. It produces two consecutive resonant frequencies with a tuning fork, which are 375 Hz and 450 Hz. The mass of the string is

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Answer: (d)

  1. Consecutive resonant frequencies differ by the fundamental: n1=450−375=75 Hzn_1 = 450 - 375 = 75\text{ Hz}.
  2. v=T/μ=1802×10−3=300 m/sv = \sqrt{T/\mu} = \sqrt{\dfrac{180}{2\times10^{-3}}} = 300\text{ m/s}
  3. L=v2n1=300150=2 mL = \dfrac{v}{2n_1} = \dfrac{300}{150} = 2\text{ m}
  4. Mass =μL=2×10−3×2=4×10−3 kg=4 g= \mu L = 2\times10^{-3}\times 2 = 4\times10^{-3}\text{ kg} = 4\text{ g}
  5. So the answer is (d) 4 gram.
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