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प्रश्न
The displacement time graph of a particle executing S.H.M. is shown in figure. Which of the following statement is/are true?
- The force is zero at `t = (T)/4`.
- The acceleration is maximum at `t = (4T)/4`.
- The velocity is maximum at `t = T/4`.
- The P.E. is equal to K.E. of oscillation at `t = T/2`.
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उत्तर
a, b and c
Explanation:
Consider the diagram
From the given diagram; it is clear that
- At `t = (3T)/4`, the displacement of the particle is zero. Hence. the particle executing SHM will be at the mean position ie., x = 0. So, acceleration is zero and force is also zero.
- At t = `(4T)/4`, displacement is maximum i.e., extreme position, so acceleration is maximum.
- At t = `T/4`, corresponds to the mean position. so velocity will be maximum at this position.
- At t = `(2T)/4 = T/2`, corresponds to extreme position, so KE = 0 and PE = maximum.
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