11 OSCILLATORY MOTION 11.6 Uniform circular motion

k

‚

.,

Worked example 11.4: Energy in simple harmonic motion

Question: A block of mass m = 4 kg is attached to a spring, and undergoes simple

harmonic motion with a period of T = 0.35 s. The total energy of the system is

E = 2.5 J. What is the force constant of the spring? What is the amplitude of the
motion?

Answer: The angular frequency of the motion is

2 π

ω = =

T

2 π (^)
0.35
= 17.95 rad./s.
Now, ω =
q
k/m for a mass on a spring. Rearrangement of this formula yields
k = m ω^2 = 4 × 17.95 × 17.95 = 1289.1 N/m.
The total energy of a system executing simple harmonic motion is E = a^2 k/2.
Rearrangement of this formula gives
a =
‚
., 2 E
(^)

2 × 2.5
1289.1
= 0.06228 m.
Thus, the force constant is 1289.1 N/m and the amplitude is 0.06228 m.
Worked example 11.5: Gravity on a new planet
Question: Having landed on a newly discovered planet, an astronaut sets up a
simple pendulum of length 0.6 m, and finds that it makes 51 complete oscillations
in 1 minute. The amplitude of the oscillations is small compared to the length of
the pendulum. What is the surface gravitational acceleration on the planet?
Answer: The frequency of the oscillations is
51
f =
60
Hence, the angular frequency is
= 0.85 Hz.
ω = 2 π f = 2 × π × 1.833 = 5.341 rad./s.