10.2. Mass on a Spring http://www.ck12.org
Frequency
Thefrequencyof periodic motion is defined as the reciprocal of the periodf=T^1. For example, if an object requires
a quarter of a second, 0.25 s, to complete one cycle of its motion, it has a frequency of
f= 0 .25 s^1 = 4. (^01) s= 4 .0 Hz.
The unit Hz isHertz, named after the German physicist Heinrich Rudolf Hertz (1857-1894), who we will discuss
again later. The Hertz is used in place of revolutions per second, as well as cycles per second. Hertz is equivalent to
1
s, cycle and revolution and has no SI units.
Check Your Understanding
A block attached to the end of a spring, displaced from equilibrium and released, has a period of 5.00 s.
a. What is the frequency of the block’s vibrations?
Answer:f=^1 T= 5 .00 s^1 = 0 .20 Hz
b. How many times has the blocked passed its equilibrium position, from the time of its release, if it is permitted to
vibrate for only 6.00 s?
Answer:The block will pass its equilibrium position twice each cycle. Since one cycle of its motion takes 5.00
s it will have completed only one cycle during 6.00 s and therefore it will pass its equilibrium position 2 times.
Regardless of where the block is released from equilibrium it will take 1.25 s to reach equilibrium (why?). Since the
block returns to its release position in 5 s, it will not have enough time to pass its equilibrium position a third time.
Amplitude
Theamplitudeof an object’s periodic motion is defined as the maximum distance from equilibrium. We usually use
the symbolAfor amplitude.
Check Your Understanding
A horizontal spring with a block attached is stretched 8.40 cm beyond its equilibrium position. What is the maximum
compression of the spring after passing through its equilibrium position?
Answer:A block initially stretched 8.40 cm beyond its equilibrium position will have amplitude 8.40 cm. Since the
motion of the block is symmetric about the equilibrium position, the spring will have a maximum compression of
8.40 cm.
Illustrative Example 1
a. A 5.00-kg block inFigure10.7 is attached to a spring with a force constant ofk= 75. 0 Nm. The spring is stretched
+0.35m beyond its equilibrium position 0 and released. Use the conservation of mechanical energy to find the
velocity of the block when it is at position 0.15 m.
Answer:
The law of conservation of mechanical energy states thatEi=Ef. Let us choose the initial state at which the block
is located at the 0.35 m position and the final state when the block is located at the 0.15 m position. At the position
0.35 m, all of the energy of the block-spring system is stored as potential energy in the spring. The initial energy of
the block-spring system isEi=^12 kA^2. In its final state, the block-spring system will have both potential and kinetic
energy since the spring is still stretched at the 0.15 m position and the block now is in motion. If the final position