11 OSCILLATORY MOTION
11.6 Uniform circular motion
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 ω= q
2π 2π = = 17.95 rad./s. T 0.35
Now, ω = k/m for a mass on a spring. Rearrangement of this formula yields k = m ω2 = 4 × 17.95 × 17.95 = 1289.1 N/m.
The total energy of a system executing simple harmonic motion is E = a2 k/2. Rearrangement of this formula gives a=
v u u2 E t
k
=
v u u2 t
× 2.5 = 0.06228 m. 1289.1
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 f=
51 = 0.85 Hz. 60
Hence, the angular frequency is ω = 2 π f = 2 × π × 1.833 = 5.341 rad./s. 251