A mass $m$ is attached to two springs as shown in the figure. The spring constants of the two springs are $K_1$ and $K_2$. For the frictionless surface, the time period of oscillation of mass $m$ is:
- \(2 \pi \sqrt{\frac{m}{K_1-K_2}}\)
- \(\frac{1}{2 \pi} \sqrt{\frac{K_1-K_2}{m}}\)
- \(\frac{1}{2 \pi} \sqrt{\frac{K_1+K_2}{m}}\)
- \(2 \pi \sqrt{\frac{m}{K_1+K_2}}\)
The Correct Option is D
Approach Solution - 1
\(\begin{aligned} & K_{e q}=K_1+K_2 \\ & T=2 \pi \sqrt{\frac{m}{K_{e q}}}=2 \pi \sqrt{\frac{m}{K_1+K_2}} \end{aligned}\)
Approach Solution -2
Springs in Parallel Oscillation Problem
Step 1: Springs in Parallel
The two springs are connected in parallel, meaning that they experience the same displacement when the mass oscillates.
Step 2: Equivalent Spring Constant
For springs in parallel, the equivalent spring constant (\( K_{eq} \)) is the sum of the individual spring constants:
\( K_{eq} = K_1 + K_2 \)
Step 3: Time Period of Oscillation
The time period (T) of oscillation for a mass-spring system is given by:
\( T = 2\pi \sqrt{\frac{m}{K}} \)
where \( m \) is the mass and \( K \) is the spring constant. In this case, we use the equivalent spring constant:
\( T = 2\pi \sqrt{\frac{m}{K_{eq}}} = 2\pi \sqrt{\frac{m}{K_1 + K_2}} \)
Conclusion:
The time period of oscillation of mass \( m \) is \( \mathbf{2\pi \sqrt{\frac{m}{K_1+K_2}}} \) (Option 4).
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