For a simple harmonic motion in a mass spring system shown, the s… — Solution | JEE Physics

MCQEasyJEE 2023Simple Harmonic Motion (SHM)

For a simple harmonic motion in a mass spring system shown, the surface is frictionless. When the mass of the block is $1 \, \text{kg}$ , the angular frequency is $\omega_1$ . When the mass block is $2 \, \text{kg}$ the angular frequency is $\omega_2$ . The ratio $\frac{\omega_2}{\omega_1}$ is:

A horizontal spring attached to a wall on the left and connected to a block on a frictionless horizontal surface.

A
$\sqrt{2}$
B
$\frac{1}{\sqrt{2}}$
C
$2$
D
$\frac{1}{2}$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: A mass-spring system on a frictionless surface. For mass $m_1 = 1 \, \text{kg}$ , angular frequency is $\omega_1$ . For mass $m_2 = 2 \, \text{kg}$ , angular frequency is $\omega_2$ .

Find: The ratio $\frac{\omega_2}{\omega_1}$ .

For a spring-mass system,

\omega = \sqrt{\frac{k}{m}}

Therefore,

\omega_1 = \sqrt{\frac{k}{m_1}}, \qquad \omega_2 = \sqrt{\frac{k}{m_2}}

Taking the ratio,

\frac{\omega_2}{\omega_1} = \sqrt{\frac{m_1}{m_2}}

Substituting $m_1 = 1$ and $m_2 = 2$ ,

\frac{\omega_2}{\omega_1} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}

Therefore, the correct option is B. The solution labels it as option D, but the listed value matches option B.

Common mistakes

Using $\omega \propto \sqrt{m}$ instead of $\omega \propto \frac{1}{\sqrt{m}}$ . This reverses the dependence on mass. For a spring-mass system, increasing mass decreases angular frequency.
Taking the ratio as $\frac{\omega_2}{\omega_1} = \sqrt{\frac{m_2}{m_1}}$ . The mass terms appear inverted because $\omega = \sqrt{\frac{k}{m}}$ , so the correct ratio is $\sqrt{\frac{m_1}{m_2}}$ .

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Answer

Step-by-step solution

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