MHT CET · Physics · Oscillations
A spring executes S.H.M. with mass \(10 \mathrm{~kg}\) attached to it. The force constant of the
spring is \(10 \mathrm{~N} / \mathrm{m}\). If at any instant its velocity is \(40 \mathrm{~cm} / \mathrm{s}\), the displacement at that
instant is (Amplitude of S.H.M. \(=0.5 \mathrm{~m}\) )
- A \(0.3 \mathrm{~m}\)
- B \(0.2 \mathrm{~m}\)
- C \(0.4 \mathrm{~m}\)
- D \(0.45 \mathrm{~m}\)
Answer & Solution
Correct Answer
(A) \(0.3 \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
V=0 \sqrt{A^{2}-x^{2}}=\sqrt{\frac{k}{m}} \cdot \sqrt{A^{2}-x^{2}} \\
k=10 \mathrm{~N} / \mathrm{m}, \mathrm{m}=10 \mathrm{~kg}, A=0.5 \mathrm{~m} \\
V=40 \mathrm{~cm} / \mathrm{s}=0.4 \mathrm{~m} / \mathrm{s}
\end{array}\)
Substituting the values and solving we get \(\mathrm{x}=0.3 \mathrm{~m}\)
V=0 \sqrt{A^{2}-x^{2}}=\sqrt{\frac{k}{m}} \cdot \sqrt{A^{2}-x^{2}} \\
k=10 \mathrm{~N} / \mathrm{m}, \mathrm{m}=10 \mathrm{~kg}, A=0.5 \mathrm{~m} \\
V=40 \mathrm{~cm} / \mathrm{s}=0.4 \mathrm{~m} / \mathrm{s}
\end{array}\)
Substituting the values and solving we get \(\mathrm{x}=0.3 \mathrm{~m}\)
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