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JEE Main 20199 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
9 April
Shift
Evening Shift
Subject
Physics
Chapter
Waves and Sound

JEE Main 2019 · 9 April · Evening Shift · Physics · Waves and Sound

A string 2.0m long and fixed at its ends is driven by a 240Hz...

A string 2.0m2.0m long and fixed at its ends is driven by a 240Hz240Hz vibrator. The string vibrates in its third harmonic mode. The speed of the wave and its fundamental frequency is

Options

  1. A320ms−1,120Hz320{m s}^{- 1}, 120Hz
  2. B320ms−1,80Hz320{m s}^{- 1}, 80Hz
  3. C180ms−1,80Hz180{m s}^{- 1}, 80Hz
  4. D180ms−1,120Hz180{m s}^{- 1}, 120Hz

Correct answer

B. 320ms−1,80Hz320{m s}^{- 1}, 80Hz

Step-by-step solution

For a string fixed at both the ends,
f=n2Lvf = \dfrac{n}{2L}v
For third harmonic, 240=32(2)v240 = \dfrac{3}{2 ( 2 )}v
⇒v=240×43⇒ v = \dfrac{240 \times 4}{3}
=320ms−1= 320m{s}^{- 1}
Fundamental frequency f0=12Lv{f}_{0} = \dfrac{1}{2L}v
=14×320=80Hz= \dfrac{1}{4} \times 320 = 80Hz
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