CK-12-Physics - Intermediate

http://www.ck12.org Chapter 10. Periodic Motion

Illustrative Example 2

A 3.50-kg block stretches a vertical spring 9.50 cm beyond its unstretched lengthL 0 (Figure10.8) as shown at
positionA. If the block is pulled down, stretching the spring an additional amountd(positionB), and released, with
what frequency will the block vibrate?

FIGURE 10.8

Answer:

At first glance it may seem that we don’t have enough information to solve the problem, since we require both the
mass of the block and the spring constant in order to find the period of the block and eventually the frequency. But
we must remember that the block is in equilibrium at positionAinFigure10.8. The Free-Body-Diagram of the
block is drawn at positionA. The weight of the blockmgis balanced by the force that the spring exerts in the upward
directionkx. The net force on the block at positionAis zero, so∑F=kx−mg= 0 →k=mgx.

We may then use the equationT= 2 π

√m

k.

We will use this equation in order to write a useful equation for the frequency:

Since the frequency isf=T^1 , substituting 2π

√m

kforTgivesf=

1

2 π

√m

k

= 21 π

√

k

m→f=

1

2 π

√

k

m

(Equation 3).

Substitutingk=mgx into Equation 3, we have

f=

1

2 π

√

g

x

=→f=

1

2 π

√

9. 81

0. 0950

= 1. 617 → 1. 62 Hz

Notice the additional distancedthe spring was stretched did not enter into the calculation for the period of frequency.