Question

A mass is suspended separately by two different springs in successive order then time periods is \(t_1\) and \(t_2\) respectively. If it is connected by both spring as shown in figure then time period is \(t_0\), the correct relation is

• A. $ t_0^2 = t_1^2 + t_2^2 $
• B. $ t_0^{ - 2} = t_1^{ -2} + t_2^{-2} $
• C. $ t_0^{ - 1} = t_1^{ - 1} + t_2^{ - 1} $
• D. $ t_0 = t_1 + t_2 $

Answer 1

The Correct Option is B

To solve this problem, we need to understand the behavior of a mass-spring system and the formulas governing the time periods of oscillation.

When a mass \( m \) is suspended by a spring with spring constant \( k \), the time period of oscillation \( T \) is given by:

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

For two springs, the time periods are given as \( t_1 \) and \( t_2 \) when the mass is attached to the springs individually. Let the spring constants of the two springs be \( k_1 \) and \( k_2 \). Thus, we have:

• t_1 = 2\pi \sqrt{\frac{m}{k_1}}
• t_2 = 2\pi \sqrt{\frac{m}{k_2}}

When the springs are connected in parallel (as shown in the image), the effective spring constant \( k_{\text{eff}} \) is the sum of the individual spring constants:

k_{\text{eff}} = k_1 + k_2

The time period for the parallel spring system, \( t_0 \), is then:

t_0 = 2\pi \sqrt{\frac{m}{k_{\text{eff}}}} = 2\pi \sqrt{\frac{m}{k_1 + k_2}}

Squaring both sides of the above three equations, we get:

• t_1^2 = 4\pi^2 \frac{m}{k_1}
• t_2^2 = 4\pi^2 \frac{m}{k_2}
• t_0^2 = 4\pi^2 \frac{m}{k_1 + k_2}

From these, we find the relationship:

\frac{1}{t_0^2} = \frac{k_1 + k_2}{4\pi^2 m} = \frac{1}{t_1^2} + \frac{1}{t_2^2}

This confirms the correct relationship as:

t_0^{-2} = t_1^{-2} + t_2^{-2}

Hence, the correct answer is:

t_0^{-2} = t_1^{-2} + t_2^{-2}