Question

An object is undergoing SMH with a period of 1.200 sec and amplitude

of 0.600 m. At t=0 the object is at x=0 and is moving in the negative x direction.

(a) What is the equation of motion that describes this objects motion?

(b) What is the value of f in the equation?

(c) When will x be zero for the second time?

(d) At what time will x be at a maximum positive magnitude?

(e) How far is the object from the equilibrium position at t=0.480 s?

Answer 1

GIVEN DATA

Time Period T = 1.2 sec

Amplitude A = 0.6 m

(a) Equation of motion x = A Sin(t) = A Sin(2f)

where is angular frequency and f is the frequency

(b) frequency f = 1/T = 1/1.2 = 0.8333 Hz

(c) X value will be zero for every half period or any SHM will pass through mean position twice in one period

i.e x will be zero second time is T/2 = 1.2/2 = 0.6 sec

(d) Intially Oscillator moved in - ve x direction, after 0.6 sec it will pass through mean postion. hence after 0.6 Oscillator will move in + x direction

(e) x =  x = A Sin(t) = 0.6 Sin()

= 0.6 Sin

= 0.6 Sin