(a) Sketch a diagram of a simple pendulum performing simple harmonic motion and indicate positions of maximum potential energy and kinetic energy.
(b) A body moving with simple harmonic motion in a straight line has velocity, v and acceleration, a, when the instantaneous displacement, x in cm, from its maximum position is given by x = 2.5 sin 0.4 \(\pi t\), where t is in seconds. Determine the magnitude of the maximum (i) veloxity; (ii) acceleration
(c) A mass m attached to a light spiral is caused to perform simple harmonic motion of frequency
f = \(\frac{1}{2 \pi} \sqrt{\frac{k}{m}}\), where k is the force constant of the spring.
(i) Explain the physical significance of \(\sqrt{\frac{k}{m}}\).
(ii) If m = 0.30 kg, k = 30Nm\(^{-1}\) and the maximum position is 0.015m, calculate the maximum;
(i) kinetic energy
(ii) tension in the spring during the motion [g = 10 ms\(^{-1}\), \(\pi\) = 3.142]
Verified Correct Answer: .
Physical Principles & Numerical Solution
(a)
(b)(i) V = rw = 2.5 x 10\(^{-2}\) x 0.4 \(\pi\)
= 3.142 x 10\(^{-2}\)m/s
(ii) a = rw\(^2\) = 2.5 x 1.0\(^{-2}\) x (0.4 \(\pi^2\))
(c) (i) \(\frac{k}{m}\) = w. H is the rate of change of angular displacement. H is measured in radians per second
(ii) (1) K.E\(_{\text{max}} \frac{1}{2}MV^2_{\text{max}} = \frac{1}{2} mr^2w^2\)
But w = \(\sqrt{\frac{k}{m}} = w^2 = \frac{k}{m} = \frac{30}{0.3} = 100 rad/s^2\)
K.E\(_{\text{max}} = \frac{1}{2}mr^2w^2 = \frac{1}{2} \times 0.30 \times (0.015)^2\) x 100
= 3.38 x 10\(^{3}\)J
T\(_{\text{max}}\) = Weight + Ke = Mg + Ke
= 0.30 x 10 + 30 x 0.015
= 3.45N
Source: Myschool Past Questions Archive (Record #53832)