Q43 / 107 · NTA Abhyas (JEE Main) 2020 · Physics · Oscillations

An elastic string has a force constant k and mass m. The string hangs vertically and a block of unknown mass is attached to its bottom end. It is known that the mass of the block is much greater than that of the string. The hanging block stretches the string the twice it's the relaxed length. How long t would it take for a low amplitude transverse pulse to travel the length of the string stretched by the hanging block? (Assume that the string follows Hook's law)

  • Correct answer
    A
    2mk
  • B
    mk
  • C
    m2k
  • D
    2m3k

Solution

Since the mass of the string is small compared to the mass m , the tension force is approximately constant

∴ T=mg

kx=mg, x=mgk

x=2L given

Where L- Relaxed length.

New length=2L

=2mgk

∴ mass per unit length

µ=mk2mg=k2g

Speed of wave V=Tμ

=mg×2mgmk

V=g2mk

time=2LV

=2mgk×1gk2m

time=2mk

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

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