Page 186 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 14] SIMPLE HARMONIC MOTION 171

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2
For a body in simple harmonic motion, a =−4π f s, which is a maximum when s =±A, where A is the
amplitude. Hence
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2
a max =±4π f A
The acceleration is negative when s =+A, and the acceleration is positive when s =−A.

SOLVED PROBLEM 14.10
Each piston of a certain car engine has a mass of 0.5 kg and has a “stroke” (total travel distance) of
120 mm. When the engine is operating at 3000 rev/min, find (a) the maximum velocity of each piston,
(b) its maximum acceleration, and (c) the maximum force on it. Assume that the pistons move up and
down in simple harmonic motion, which is approximately true.
(a) The frequency of oscillation of each piston is 3000 cycles/min, which is
3000 cycles/min
f = = 50 Hz
60 s/min
The amplitude of the motion is half the stroke, so A = 60 mm = 0.6 m. Hence the maximum piston velocity is
−1
v max = 2πfA = (2π)(50 s )(0.06 m) = 18.8 m/s
(b) The maximum acceleration is
3
2
−1 2
2
2
a max = 4π f A = (4π )(50 s ) (0.06 m) = 5.92 × 10 m/s
(c) The maximum force on each piston is
3
2
3
F max = ma max = (0.5kg)(5.92 × 10 m/s ) = 2.96 × 10 N = 2.96 kN
ENERGY

The energy of an oscillator shifts back and forth between its kinetic and potential forms. To find the total energy,
we note from Prob. 14.8 that the maximum velocity of an oscillating body is v max = 2πfA and it occurs when
the displacement is s = 0. When s = 0, the PE of the body is 0 and its kinetic energy is a maximum, where
1
2
2
KE max = mv 2 = 2π f A 2
2 max
Hence the total energy E of the oscillator is
2
2
E = 2π f A 2
At s = A and s =−A, the extremes of the motion, the body is momentarily at rest and all its energy is PE. At
s = 0, as noted, all the energy is KE. At all other values of s the total energy is divided between KE and PE.

SOLVED PROBLEM 14.11

The amplitude of a simple harmonic oscillator is doubled. How does this affect (a) the period, (b) the
total energy, and (c) the maximum velocity of the oscillator?

(a) The period of such an oscillator does not depend on the amplitude of its motion; hence T is unchanged.
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(b) The total energy of the oscillator is equal to the elastic potential energy kA at either extreme of its motion,
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when v = 0. Hence doubling A means that the total energy increases fourfold.
(c) The maximum velocity occurs at s = 0, the equilibrium position, when the entire energy of the oscillator is KE.
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2
Since KE = mv and the total energy is four times greater than originally, v max must double.
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