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Question:
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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. $=0.5 \mathrm{~m}$ )
Options:
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. $=0.5 \mathrm{~m}$ )
Solution:
1361 Upvotes
Verified Answer
The correct answer is:
$0.3 \mathrm{~m}$
$\begin{array}{l}
V=0 \sqrt{A^{2}-x^{2}}=\sqrt{\frac{k}{m}} \cdot \sqrt{A^{2}-x^{2}} \\
k=10 \mathrm{~N} / \mathrm{m}, \mathrm{m}=10 \mathrm{~kg}, A=0.5 \mathrm{~m} \\
V=40 \mathrm{~cm} / \mathrm{s}=0.4 \mathrm{~m} / \mathrm{s}
\end{array}$
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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