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Standard XII

Physics

SHM expression

A block perfo...

Question

A block performs simple harmonic motion with equilibrium point $x = 0$ . Graph of acceleration of the block as a function of time is shown. Which of the following statement is correct about the block?

Displacement from equilibrium is maximum at

t = 4 s

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Speed is maximum at

t = 4 s

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Speed is maximum at

t = 2 s

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Speed is maximum at

t = 3 s

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Solution

The correct option is A Speed is maximum at $t = 4 s$
From the figure given in the question, the speed of block will be maximum at $t = 0, 2$ and $4$ . So, the block can execute the simple harmonic motion. At extreme points at time $t = 1 s$ and $t = 3 s$ , the direction of acceleration is change. So, at $t = 4 s$ the speed of the block is maximum.

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A block performs simple harmonic motion with equilibrium point x=0. Graph of acceleration of the block as a function of time is shown. Which of the following statement is correct about the block?

A block performs simple harmonic motion with equilibrium point $x = 0$ . Graph of acceleration of the block as a function of time is shown. Which of the following statement is correct about the block?