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Question

The x−t graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at t=43 s is


A
3π232 cm/s2
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B
π232 cm/s2
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C
π232 cm/s2
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D
3π232 cm/s2
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Solution

The correct option is D 3π232 cm/s2
From the xt graph of the particle,


At, t=0, the particle is at x=0.
Hence, the particle is starting SHM from its mean position.
The equation of SHM of particle will be,
x=Asinωt ...........(1)
From the graph, one oscillation has been completed in time, T=8 s and maximum displacement, A=1 cm
So, angular frequency, ω=2πT=2π8=π4 rad/s
On putting above data in (1)
x=1sin(πt4)
At, t=43 s
x=1sin(π4×43)=32 cm
Now,
Acceleration of the particle at t=43 s,
a=ω2x=(π4)2×32
a=3π232 cm/s2

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