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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Simple Harmonic Motion

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Simple Harmonic Motion

Determine the time period of the mass m , assuming the string,...

Determine the time period of the mass mm, assuming the string, spring, and pulley shown in the figure are light.
Determine the time period of the mass m , assuming the string, spring, and pulley shown in the figure are light.

Options

  1. Aπmk\pi \sqrt{\dfrac{m}{k}}
  2. B2πm2k2\pi \sqrt{\dfrac{m}{2k}}
  3. C2πmk2\pi \sqrt{\dfrac{m}{k}}
  4. D2π2mk2\pi \sqrt{\dfrac{2m}{k}}

Correct answer

C. 2πmk2\pi \sqrt{\dfrac{m}{k}}

Step-by-step solution

Let the equilibrium extension of the spring be x0x_0.

At equilibrium, the tension in the string balances the weight of the mass mm.

T0=mgT_0 = mg

The tension is also equal to the spring force.

T0=kx0⇒kx0=mgT_0 = kx_0 \Rightarrow kx_0 = mg

When the mass mm is displaced downwards by a distance xx, the string pulls the spring upwards by the same distance xx.

The new extension of the spring becomes x0+xx_0 + x.

The new tension in the string is T=k(x0+x)T = k(x_0 + x).

The net restoring force acting on the mass mm in the downward direction is:

F=mg−TF = mg - T

F=mg−k(x0+x)F = mg - k(x_0 + x)

Substituting mg=kx0mg = kx_0:

F=kx0−kx0−kx=−kxF = kx_0 - kx_0 - kx = -kx

The acceleration of the mass is a=Fm=−kmxa = \dfrac{F}{m} = -\dfrac{k}{m}x.

This is the equation of a simple harmonic motion a=−ω2xa = -\omega^2 x, where ω2=km\omega^2 = \dfrac{k}{m}.

The time period of oscillation is given by T=2πωT = \dfrac{2\pi}{\omega}.

T=2πmkT = 2\pi \sqrt{\dfrac{m}{k}}
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