Question

A mass \(m\) attached to free end of a spring executes SHM with a period of \(1 \,s\) If the mass is increased by \(3\, kg\) the period of oscillation increases by one second, the value of mass \(m\) is ______\(kg.\)

Hint:

The time period of SHM depends on mass and spring constant: \(T=2\pi\sqrt{\frac{m}{k}}\). For mass changes, compare squared ratios of T.

Answer 1

Correct Answer: 1

To solve this problem, we need to use the formula for the period of simple harmonic motion (SHM) of a mass-spring system: \(T = 2\pi \sqrt{\frac{m}{k}}\), where \(T\) is the period, \(m\) is the mass, and \(k\) is the spring constant.

Initially, the period \(T_1 = 1 \, \text{s}\), so:

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

Squaring both sides, we get:

\(\frac{m}{k} = \frac{1}{4\pi^2}\)

When the mass is increased by 3 kg, the new period \(T_2 = 2 \, \text{s}\):

\(2 = 2\pi \sqrt{\frac{m+3}{k}}\)

Squaring both sides, we have:

\(\frac{m+3}{k} = \frac{1}{\pi^2}\)

Now, we have two equations:

(1) \(\frac{m}{k} = \frac{1}{4\pi^2}\)

(2) \(\frac{m+3}{k} = \frac{1}{\pi^2}\)

Subtract equation (1) from equation (2):

\(\frac{3}{k} = \frac{1}{\pi^2} - \frac{1}{4\pi^2}\)

\(\frac{3}{k} = \frac{4 - 1}{4\pi^2}\)

\(\frac{3}{k} = \frac{3}{4\pi^2}\)

Solving for \(k\), we get:

\(k = 4\pi^2\)

Substitute \(k = 4\pi^2\) back into equation (1):

\(\frac{m}{4\pi^2} = \frac{1}{4\pi^2}\)

From which, \(m = 1\) kg.

Thus, the value of the mass \(m\) is \(1\) kg, which is within the given range of 1 to 1.