Question

A mass m is suspended from the two coupled springs connected in series. The force constant for springs are $ k_1 $ and $ k_2$. The time period of the suspended mass will be

• A. T = 2 $ \pi \sqrt{ \frac{ m}{ k_1 - k_2 }} $
• B. T = 2 $ \pi \sqrt{ \frac{ m k_1 k_2}{ k_1 + k_2 }} $
• C. T = 2 $ \pi \sqrt{ \frac{ m}{ k_1 + k_2 }} $
• D. T = 2 $ \pi \sqrt{ \frac{ m ( k_1 + k_2)}{ k_1k_2 }} $

Answer 1

The Correct Option is D

To determine the time period of a mass $ m $ suspended from two springs connected in series with force constants $ k_1 $ and $ k_2 $, we can use the formula for equivalent spring constant $ k_{\text{eq}} $ of springs in series:

\(\frac{1}{k_{\text{eq}}} = \frac{1}{k_1} + \frac{1}{k_2}\)

Re-arranging the equation, we get:

k_{\text{eq}} = \frac{k_1 k_2}{k_1 + k_2}

This means the system can be treated as a single spring with spring constant $ k_{\text{eq}} $. The formula for the time period $ T $ of a mass $ m $ oscillating on a spring is:

T = 2 \pi \sqrt{\frac{m}{k_{\text{eq}}}}

Substituting the expression for $ k_{\text{eq}} $, we get:

T = 2 \pi \sqrt{\frac{m}{\frac{k_1 k_2}{k_1 + k_2}}} = 2 \pi \sqrt{\frac{m (k_1 + k_2)}{k_1 k_2}}

Therefore, the correct expression for the time period of the suspended mass is:

T = 2 \pi \sqrt{\frac{m (k_1 + k_2)}{k_1 k_2}}

This matches the correct answer option provided.