Question

A 30.0 kg block at rest on a horizontal frictionless air track is connected to the wall via a spring. The equilibrium position of the mass is defined to be at x=0. Somebody pushes the mass to the position x= 0.350 m, then lets go. The mass undergoes simple harmonic motion with a period of 4.70 s. What is the position of the mass 3.713 s after the mass is released?

Consider the same mass and spring discussed in the previous problem. What is the magnitude of the maximum acceleration the mass undergoes during its motion?

Answer 1

Given,

Mass (m) =30 kg

Maximum amplitude (a)=0.350m

Time period (T)=4.7 sec

Equation of simple harmonic motion,

x=a sin(t+)

Angular velocity ()=2/T

=2*3.14/4.7

=1.336rad/sec

x=asin(t) initial phase =0

Substituting the values

x=0.35sin(1.336*3.713)

x=0.03026m from equilibrium position

Acceleration A= - 2x

Maximum accelerationis when x= a( maximum amplitude)

Maximum acceleration=- 2a

= - 1.336*1.336*0.35

=.6247m/sec (taking magnitude only)