JEE Mains · Physics · STD 11 - Oscillations
- JEE Mains 2020
- Hard
- MCQ
The displacement time graph of a particle executing \(S.H.M.\) is given in figure: (sketch is schematic and not to scale) Which of the following statements is are true for this motion? \((A)\) The force is zero \(t=\frac{3 T}{4}\) \((B)\) The acceleration is maximum at \(t=T\) \((C)\) The speed is maximum at \(t =\frac{ T }{4}\) \((D)\) The \(P.E.\) is equal to \(K.E.\) of the oscillation at \(t=\frac{T}{2}\)

Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(( A ),( B )\) and \(( C )\)
Step-by-step Solution
Detailed explanation
Sol. \((A)\) \(F =\) ma \(\quad a =-\omega^{2} x\)
at \(\frac{3 T }{4}\) displacement zero \(( x =0),\) so \(a =0\)
\(F=0\)
\((B)\) at \(t=T \quad\) displacement \((x)=A\) \(x\) maximum, So acceleration is maximum.
\((C)\) \(V =\omega \sqrt{ A ^{2}- x ^{2}}\)
\(V _{\max }\) at \(x =0\)
\(V _{\max }= A \omega\)
at \(t =\frac{ T }{4}, x =0, \quad\) So \(V _{ max }\)
\((D)\) \(KE = PE\)
\(\therefore\) at \(x=\frac{A}{\sqrt{2}}\)
at \(t=\frac{T}{2} \quad x=-A \quad\) (So not possible)
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