JEE Mains · Physics · STD 11 - 13. oscillations
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 \(t=\frac{T}{2}\)
- A \(( A ),( B )\) and \(( D )\)
- B \(( B ),( C )\) and \(( D )\)
- C \((A)\) and \((D)\)
- D \(( A ),( B )\) and \(( C )\)
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…
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