Question

A mass \( m \) hanging from a spring is doing simple harmonic motion with frequency \( f \). If the mass is increased by 4 times, then frequency will be:

• A. \(2f\)
• B. \( \frac{f}{2} \)
• C. \(4f\)
• D. \( \frac{f}{4} \)

Hint:

Frequency in SHM is inversely proportional to square root of mass.

Answer 1

The Correct Option is B

The frequency of oscillation for a mass-spring system in simple harmonic motion is given by the formula:

\(f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}\)

where:

• \(f\) is the frequency of oscillation,
• \(k\) is the spring constant,
• \(m\) is the mass.

When the mass \(m\) is increased by 4 times, the new mass becomes \(4m\).

The new frequency \(f'\) is then given by:

\(f' = \frac{1}{2\pi} \sqrt{\frac{k}{4m}}\)

We can rewrite this equation as:

\(f' = \frac{1}{2\pi} \frac{1}{\sqrt{4}} \sqrt{\frac{k}{m}} = \frac{1}{2} \cdot \frac{1}{2\pi} \sqrt{\frac{k}{m}}\)

Since the original frequency \(f\) is:

\(f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}\)

The new frequency \(f'\) is:

\(f' = \frac{f}{2}\)

Hence, when the mass is increased by 4 times, the frequency of the oscillations reduces to half of the original frequency.

Therefore, the correct answer is \(\frac{f}{2}\).

This explains why the given option \(\frac{f}{2}\) is correct.