Question

a body executes simple harmonic motion along x axis its displacement varies with time according to the equation X =1.2 cos( 50pie t) where X is in metre t is in seconds find the amplitude and the time period of the particle​

Answer 1

Explanation:

The given equation of simple harmonic motion is

x(t)=5cos[2πt+

4

π

]

Compare the given equation with standard equation of SHM

x(t)=Acos(ωt+ϕ)

we get, ω=2πs

−1

At t=1.5s

Displacement, x(t)=5cos[2π×1.5+

4

π

]=5cos[3π+

4

π

]=−5cos[

4

π

]=−5×0.707m=−3.54m

Acceleration, a=−ω

2

×displacement=−(2πs

−1

)

2

×(−3.54m)=140ms

−2