Question #273503

A particle moving along the x โ€“ axis in simple harmonic motion starts from the equilibrium position, the origin, at time t = 0 and moves to the right. The position of the particle after time, t, is given by the following equation: ๐‘ฅ = 2.0 sin 3๐œ‹๐‘ก m Calculate:

i. The maximum velocity of the motion.

ii. The acceleration of the object after 2 seconds (i.e. at t = 2 s).

iii. The distance travelled by the particle from t = 0 s to t = 2 s.


1
Expert's answer
2021-11-30T11:19:22-0500

Maximum velocity:


vmax=2โ‹…3ฯ€=6ฯ€ m/s.v_\text{max}=2ยท3\pi=6\pi\text{ m/s}.

The acceleration:


a=x"(t)=โˆ’18ฯ€2sinโก(3ฯ€t)=โˆ’57 m/s2.a=x"(t)=-18\pi^2\sin(3\pi t)=-57\text{ m/s}^2.

The distance:


d=โˆซ02xโ€ฒ(t)dt=0.d=\int^2_0x'(t)\text dt=0.


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