Question

A mass '$m_1$' is suspended from a spring of negligible mass. A spring is pulled slightly in downward direction and released, mass performs S.H.M. of period '$T_1$'. If the mass is increased by '$m_2$', the time period becomes '$T_2$'. The ratio $\frac{m_2}{m_1}$ is

• A. $\frac{T_1^2 + T_2^2}{T_1^2}$
• B. $\frac{T_1 - T_2}{T_1}$
• C. $\frac{T_2^2 - T_1^2}{T_1^2}$
• D. $\frac{T_1^2 - T_2^2}{T_1^2}$

Hint:

Since $T^2 \propto m$, any fractional change in mass matches the fractional change in $T^2$. The fractional increase in mass is exactly $\frac{\Delta m}{m_{\text{initial}}} = \frac{m_2}{m_1}$, so it must equal the fractional change in the squared time period: $\frac{T_2^2 - T_1^2}{T_1^2}$. This bypasses setting up full proportional equations!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem describes a mass-spring system performing Simple Harmonic Motion (SHM). We need to determine how the time period changes when an additional mass is attached to the spring, and find the ratio of the added mass to the initial mass.

Step 2: Key Formula or Approach:
The time period $T$ of a mass-spring system undergoing SHM is given by the formula: $$ T = 2\pi\sqrt{\frac{m}{k}} $$ where $m$ is the total suspended mass and $k$ is the spring constant. This shows that the time period is directly proportional to the square root of the mass ($T \propto \sqrt{m}$), or equivalently, $T^2 \propto m$.

Step 3: Detailed Explanation:
Let's write down the time period equations for both scenarios:

• For the initial mass $m_1$: $$ T_1 = 2\pi\sqrt{\frac{m_1}{k}} \implies T_1^2 = \frac{4\pi^2 m_1}{k} \quad \text{--- (Equation 1)} $$

• When the mass is increased by $m_2$, the total mass becomes $(m_1 + m_2)$: $$ T_2 = 2\pi\sqrt{\frac{m_1 + m_2}{k}} \implies T_2^2 = \frac{4\pi^2 (m_1 + m_2)}{k} \quad \text{--- (Equation 2)} $$
Dividing Equation 2 by Equation 1 to eliminate the constants $4\pi^2$ and $k$: $$ \frac{T_2^2}{T_1^2} = \frac{m_1 + m_2}{m_1} $$ Split the fraction on the right-hand side: $$ \frac{T_2^2}{T_1^2} = 1 + \frac{m_2}{m_1} $$ Isolating our target ratio $\frac{m_2}{m_1}$ by subtracting 1 from both sides: $$ \frac{m_2}{m_1} = \frac{T_2^2}{T_1^2} - 1 $$ Taking a common denominator gives: $$ \frac{m_2}{m_1} = \frac{T_2^2 - T_1^2}{T_1^2} $$

Step 4: Final Answer:
The ratio of the masses is $\frac{T_2^2 - T_1^2}{T_1^2}$, which corresponds to option (C).