JEE Mains · Physics · STD 11 - Oscillations
- JEE Mains 2021
- Hard
- Numerical
A mass of \(5\, {kg}\) is connected to a spring. The potential energy curve of the simple harmonic motion executed by the system is shown in the figure. A simple pendulum of length \(4\, {m}\) has the same period of oscillation as the spring system. What is the value of acceleration due to gravity on the planet where these experiments are performed? (In \({m} / {s}^{2}\))
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(4\)
Step-by-step Solution
Detailed explanation
From potential energy curve
\({U}_{\max }=\frac{1}{2} {kA}^{2} \Rightarrow 10=\frac{1}{2} {k}(2)^{2}\)
\(\Rightarrow {k}=5\)
Now \({T}_{\text {spring }}={T}_{\text {pendulum }}\)
\(2 \pi \sqrt{\frac{5}{5}}=2 \pi \sqrt{\frac{4}{g}}\)
\(\Rightarrow 1=\sqrt{\frac{4}{g}} \Rightarrow g=4 \text { on planet }\)
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