The equation of a particle executing SHM is given by x=150sin(π3t+π2), where the amplitude is in metres and angular frequency is in rad/s. Find the maximum velocity (vmax) and maximum acceleration (amax) attained by the particle.
A
vmax=50πm/s,amax=50π23m/s2
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B
vmax=100πm/s,amax=50π23m/s2
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C
vmax=50πm/s,amax=100π23m/s2
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D
None of the above
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Solution
The correct option is Avmax=50πm/s,amax=50π23m/s2 Given, x=150sin(π3t+π2)
Comparing it with x=Asin(ωt+ϕ), we get A=150m,ω=π3rad/sec,ϕ=π2rad Vmax=Aω=150×π3=50πm/s amax=ω2A=(π3)2×150=π29×150=50π23m/s2
Thus, option (a) is the correct answer.