y=2(cm)sin[πt2+ϕ] what is the maximum acceleration of the particle doing the S.H.M ?
π2cm/s2
π22cm/s2
π4cm/s2
π24cm/s2
A
π22cm/s2
B
π2cm/s2
C
π24cm/s2
D
π4cm/s2
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Solution
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y=2sin(πt2+ϕ) Velocity of particle dydt=2×π2cos(πt2+ϕ) Acceleration d2ydt=−π242sin(πt2+ϕ) Thus amax=π22
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