Set 9 Problem number 7

• Problem

• Solution

• Explanation in terms of Figure(s), Extension

• Figure(s)

Problem

A mass of 5.5 kg is suspended from a spring with force constant 83 N / m.

• If the mass is released from rest at a point .2 meters from equilibrium, what will be the angular frequency of its motion (angular frequency is the angular velocity of the point on the reference circle which models the motion)?
• What is the average speed of the object between t=-.001 sec and t = .001 sec?
• How are the two velocities related and why?

Solution

The angular velocity of the reference point is

• angular velocity = `omega = `sqrt(k/m) = `sqrt( 83 N/m / ( 5.5 kg) ) = 3.884 radians / second.

At this angular velocity, a point on the circle will be moving at

ref circle velocity = 3.884 radians/second ( .2 meters) = .7768 meters / second

(note that on a circle of radius .2 meters, each radian corresponds to a distance of .2 meters, so that 3.884 radians corresponds to 3.884 meters).

Using the function y = A sin(`omega t) = .2 sin( 3.884 * t) to calculate the positions at the t = -.001 and t = .001 clock times, we obtain

• y(-.001 s) = -.0007769 m and y(.001 s) = .0007769 m.

Thus the average velocity between these clock times is

vAve = ( .0007769 m - -.0007769 m) / (.002 s) = 0 meters/second.

The near equality of these two velocities should not be surprising:

• As the object moves vertically true equilibrium, the point on the reference circle moves vertically through the x axis, parallel to the object.
• At this instant of object and reference point will be moving in parallel directions at the same speed.

Explanation in terms of Figure(s), Extension

The figure below depicts a reference circle and the object whose motion it models. The red reference point is at the x axis when the green object is at its equilibrium position. In this position both objects are moving in the y direction and their velocities must match.

Figure(s)