Question

Example 5.6: The maximum speed of a particle performing linear S.H.M is 0.08 m/s. If its maximum acceleration is 0.32 m/ s, calculate its (i) period and (ii) amplitude,​

Answer 1

Answer:

For a particle performing SHM, speed is given by ω√A² - x²

where ω= angular velocity/ angular frequency, A= amplitude, x= displacement ( may or may not be equal to amplitude )

When displacement (x)= 0, the particle is at its mean position and velocity is maximum.

V (max)= ωA

Similarly, the particle has acceleration (a) = ω²x

(where symbols have the same meanings )

At extreme position displacement= amplitude and acceleration is maximum

⇒ a (max)= ω²A

Now to solve the required questions, we can take the ratio of a (max) to v(max).

a/v =ω²A/ ωA = ω

⇒ ω= 0.32/ 0.08 = 4 rad/s

T = 2π/ ω = 2π/ 4= π/2 s (where π= 3.14)

Now that we have ω, let us just substitute it in either the velocity or acceleration max. equation and we will get the required amplitude.

A = v(max)/ ω = 0.08/4 = 0.02 m