Set 9 Problem number 7

• Problem

• Solution

• Explanation in terms of Figure(s), Extension

• Figure(s)

Problem

A mass of 8.5 kg released from rest at a distance of .51 meters from its equilibrium point.  If the object is subjected to a net restoring force with force constant 22 N / m, then

• Find the angular frequency of its motion (angular frequency is the angular velocity of the point on the reference circle which models the motion).
• Find 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( 22 N/m / ( 8.5 kg) ) = 1.608 radians / second.

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

ref circle velocity = 1.608 radians/second ( .51 meters) = .82 meters / second

(note that on a circle of radius .51 meters, each radian corresponds to a distance of .51 meters, so that 1.608 radians corresponds to 1.608 meters).

Using the function y = A sin(`omega t) = .51 sin( 1.608 * t) to calculate the positions at the t = -.001 and t = .001 clock times, we obtain

• y(-.001 s) = -.0008201 m and y(.001 s) = .0008201 m.

Thus the average velocity between these clock times is

vAve = ( .0008201 m - -.0008201 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)