Question

Object A is attached to ideal spring A and is moving in simple harmonic motion. Object B is attached to ideal string B and is moving in simple harmonic motion. The period and the amplitude of object B are both 2 times the corresponding values for object A. How do the maximum speeds of the two objects compare?

Answer 1

Answer:

Explanation:

Let the time period of A and B be Ta and Tb respectively. Similarly , amplitude of A and B be Aa and Ab respectively.

Given ,

Tb = 2 Ta

Ab = 2 Aa

maximum speed of a particle in SHM = ω A , where ω is angular velocity and A is amplitude .

ω = 2π / T

If maximum speed of A be Va

Va = ω A

= 2π( Aa / Ta )  

If maximum speed of B be Vb

Vb = ω A

= 2π (Ab / Tb)

Va / Vb = (2π Aa / Ta) x (Tb / 2π Ab)

= (Aa / Ab) x (Tb / T a )

=  .5 x 2 = 1

Va = Vb

Their maximum velocities will be equal.