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Q.

A 0.10 kg block oscillates back and forth along a horizontal surface. Its displacement from the origin is given by: x=(10cm)cos[(10rad/s)t+π/2rad]. What is the maximum acceleration experienced by the block 

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a

10π2m/s2

b

10π3m/s2

c

10  m/s2

d

10πm/s2

answer is A.

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Detailed Solution

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a=10×102m and ω=10rad/sec

Amax=ω2a=10×102×102=10m/sec2

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A 0.10 kg block oscillates back and forth along a horizontal surface. Its displacement from the origin is given by: x=(10 cm)cos[(10 rad​/​s) t+π​/​2 rad]. What is the maximum acceleration experienced by the block