11 OSCILLATORY MOTION 11.6 Uniform circular motion
equilibrium position, the velocity of the block is 0.75 m/s. What is the period of
oscillation of the block?
Answer: The equation of simple harmonic motion is
x = a cos(ω t − φ),where x is the displacement, and a is the amplitude. We are told that a = 0.25 m.
The velocity of the block is obtained by taking the time derivative of the above
expression:
x ̇ = −a ω sin(ω t − φ).We are told that at t = 0 (say), x = 0.1 5 m and x ̇ = 0.7 5 m/s. Hence,
0.15 = 0.25 cos(φ),
0.75 = 0.25 ω sin(φ).The first equation gives φ = cos−^1 (0.15/0.25) = 53.13◦. The second equation
yields
0.75
ω
0.2 5 × sin(53.1 3 ◦)
= 3.75 rad./s.Hence, the period of the motion is
2 π
T =
ω= 1.676 s.Worked example 11.3: Block and two springs
Question: A block of mass m = 3 kg is attached to two springs, as shown below,
and slides over a horizontal frictionless surface. Given that the force constants
of the two springs are k 1 = 1200 N/m and k 2 = 400 N/m, find the period of
oscillation of the system.
Answer: Let x 1 and x 2 represent the extensions of the first and second springs,
respectively. The net displacement x of the mass from its equilibrium position is
then given by
x = x 1 + x 2.=