Q.

What is the maximum acceleration of the particle doing the SHM y=2sinπt2+φ where 2 is in cm

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a

π2cm/s2

b

π22cm/s2

c

π4cm/s2

d

=π28cm/s2

answer is B.

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Detailed Solution

Comparing given equation with standard equation,  y=Asin(ωt+φ), we get, A=2 cm, ω=π2∴amax=ω2A =π22×2 =π22cm/s2
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What is the maximum acceleration of the particle doing the SHM y=2sinπt2+φ where 2 is in cm