Set 9 Problem number 8

• Problem

• Solution

• Generalized Solution

• Explanation in terms of Figure(s), Extension

• Figure(s)

Problem

A mass of 3.1 kg is suspended from a spring with force constant 100 N / m.

• If the mass is released from rest at a point .47 meters from equilibrium, what will be the centripetal acceleration, in m/s ^ 2, of the reference point on the circle which models the resulting simple harmonic motion?
• What acceleration would you expect for the object at the instant it reaches an extreme point?
• Explain in your summary why you would expect the object itself to undergo this acceleration at the extreme positions in its cycle.

Solution

The angular velocity of the reference point is

• angular velocity = `omega = `sqrt(k/m) = 5.679 radians / second.

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

• speed of reference point on circle = 5.679 radians/second ( .47 meters) = 2.669 meters/second.

At this velocity, the centripetal acceleration of a point on a circle of radius .47 meters will be

• centripetal acceleration = ( 2.669 meters/second) ^ 2/( .47 meters) = 15.15651 meters/second ^ 2.

Since the centripetal acceleration is directed vertically at the instant the object passes through its equilibrium point, the acceleration of the object, which has the same vertical motion as the reference point, must have the same acceleration.

In any event, the acceleration of the object can be found from Newton's Second Law as a = F/m, where F is the force at the extreme position and m the mass of the object:

• At the extreme position, .47 meters from equilibrium, the force is

• max force = ( .47 m)( 100 N/m) = 47 Newtons.

• Dividing this by the 3.1 kilogram mass we obtain the same result as above,

• force = 47 Newtons / ( 3.1 kg) = 15.15651 m/s ^ 2.

Generalized Solution

In general the centripetal acceleration of the point on the reference circle is a = v^2 / r; since v = r `omega = A `omega, we have

a = (A `omega) ^ 2 / A = A `omega^2.

The maximum force experienced by the object is Fmax = k * yMax = k * A, so the maximum acceleration is

• aMax = Fmax / mass = k * A / mass.

Since `omega = `sqrt(k / mass), k / mass = `omega^2 and the above expression becomes

• aMax = A * `omega^2,

which agrees with the expression obtained previously.

Since the acceleration function associated with position function y = A sin (`omega t) is y = - `omega^2 A sin(`omega t), we see that the maximum magnitude of the acceleration (which occurs when the value of the sine function is -1) is

aMax = `omega^2 * A,

the same as the centripetal acceleration on the reference circle.

Explanation in terms of Figure(s), Extension

The figure shows the green object at its point of maximum force, and therefore maximum acceleration, at an extreme point of its motion. Since the acceleration of the point on the reference circle is in the same direction as the acceleration of the object whose position follows the y coordinate of the reference point, the two accelerations are equal.

Figure(s)