A particle execute SHM from the mean position. its amplitude is A its time period T. At what displacement its speed is half of its maximum speed?
√3A2
√23A
2A√3
3A√2
A
√3A2
B
3A√2
C
√23A
D
2A√3
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Solution
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Velocity of simple harmonic motion:
v=ω√A2−y2
Vmax=ωA
Given that:
V=Vmax2
ω√A2−y2=ωA2
√A2−y2=A2
A2−y2=A24
y2=A2−A24=3A24
y=√32A
Hence,
option (A) is correct answer.
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