Answer to Question #157962 in Civil and Environmental Engineering for Bonevie tutor

Question

SIMPLE HARMONIC MOTION

A weight is suspended from a spring and is moving up and down in a simple

harmonic motion. At start, the weight is push up 6 m above the resting

position, and the released. After 14 seconds, the weight reaches again to its

highest position. Find the equation of the motion, and locate the weight with

respect to the resting position after 20 seconds since it was released.​

Answer 1

The equation describing the simple harmonic motion of the weight attached to the spring is

S(t) = 5 Cos t where s(t) is in centimetres.

The standard equation describing the simple harmonic motion, S(t) = A Cos ωt

We get by comparing the two equations that,

Amplitude, A = 5 cm = 0.05 m

Angular frequency, ω = 1 rad/s

At equilibrium, S(t) = 0.

(a)

At, t = 0 s

S(t) = 5 Cos t = 5 Cos 0 = 5 x 1 = 5 cm = 0.05 m

The distance of the weight at t= 0 is S(t) = 0.05m

The position is below the equilibrium position as the distance found to be positive.

(b)

Velocity of the body in simple harmonic motion, V(t) = d S(t)/dt = - 5 Sin t

At t = π/6,

V(t) = - 5 Sin π/6 = -5 x ½ = - 2.5 m/s

Velocity, V(t) = - 2.5 m/s

The weight is moving up at the given time as the velocity is negative.

(c )

At equilibrium, S(t) = 0

S(t) = 5 Cos t

0 = 5 Cos t

Cos t = 0

Time, t = π/2 s

The weight will pass the equilibrium for the first time in t = π/2 s

The speed of the particle is maximum at the equilibrium position and it is given by the relation,

V = Aω = 0.05 x 1 = 0.05 m/s

Speed of the weight at the equilibrium, V = 0.05 m/s