Question

A body of mass $m$ is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass when the mass $m$ is slightly pulled down and released, it oscillates with a time period of $3\,s$. When the mass $m$ is increased by $1\,kg$, the time period of oscillations becomes $5\, s$. The value of $m$ in $kg$ is

• A. $\frac{3}{4}$
• B. $\frac{4}{3}$
• C. $\frac{16}{9}$
• D. $\frac{9}{16}$

Answer 1

The Correct Option is D

The problem involves a spring-mass system where the time period of oscillation is given by the formula:

T = 2\pi \sqrt{\frac{m}{k}},

where T is the time period, m is the mass attached to the spring, and k is the spring constant.

Given:

• Initial mass m leads to a time period of 3\, s.
• Increased mass m + 1\text{ kg} leads to a time period of 5\, s.

Using the formula for time period, we have:

For the first scenario: \quad 3 = 2\pi \sqrt{\frac{m}{k}}

Squaring both sides:

9 = 4\pi^2 \frac{m}{k} \quad \Rightarrow \quad \frac{m}{k} = \frac{9}{4\pi^2} [Equation 1]

For the second scenario: \quad 5 = 2\pi \sqrt{\frac{m + 1}{k}}

Squaring both sides:

25 = 4\pi^2 \frac{m+1}{k} \quad \Rightarrow \quad \frac{m+1}{k} = \frac{25}{4\pi^2} [Equation 2]

From Equation 1 and Equation 2, equate the expressions for \frac{m}{k} and \frac{m+1}{k}:

\frac{9}{4\pi^2} and \frac{25}{4\pi^2}

Setting them equal as follows:

\frac{m+1}{m} = \frac{25}{9}

Rearrange the equation to solve for m:

(m+1) = m \cdot \frac{25}{9}

9m + 9 = 25m

16m = 9

m = \frac{9}{16}

Thus, the correct mass m that results in these given conditions is \frac{9}{16} \text{ kg}, which matches the provided correct answer.