(a) Sketch a diagram of a simple pendulum performing simple harmonic motion and indicate positions of maximum potential energy and kinetic energy. (b) A bod...
Question 1 Report
(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]
Answer Details
(a) The sketch below shows the energy distribution of a simple pendulum during simple harmonic motion.
At each extreme position, the bob is momentarily at rest, so its potential energy is maximum and its kinetic energy is zero. At the mean (lowest) position, its speed is greatest, so its kinetic energy is maximum and its potential energy is minimum.
(b) Comparing
\[x=2.5\sin(0.4\pi t)\ \text{cm}\]
with \(x=A\sin \omega t\),
\[A=2.5\ \text{cm}=2.5\times10^{-2}\ \text{m},\qquad \omega=0.4\pi\ \text{rad s}^{-1}.\]
(i) Maximum velocity
\[v_{\max}=A\omega\]
\[=2.5\times10^{-2}\times0.4\pi\]
\[=3.142\times10^{-2}\ \text{m s}^{-1}.\]
(ii) Maximum acceleration
\[a_{\max}=A\omega^2\]
\[=2.5\times10^{-2}(0.4\pi)^2\]
\[=3.95\times10^{-2}\ \text{m s}^{-2}.\]
(c)(i)
\[\sqrt{\frac{k}{m}}=\omega\]
is the angular frequency of the oscillation. It is the rate of change of phase (angular displacement) and is measured in \(\text{rad s}^{-1}\).
(c)(ii)
Given \(m=0.30\ \text{kg}\), \(k=30\ \text{N m}^{-1}\), \(A=0.015\ \text{m}\),
\[\omega^2=\frac{k}{m}=\frac{30}{0.30}=100\ \text{s}^{-2}.\]
(i) Maximum kinetic energy
\[K_{\max}=\frac{1}{2}mA^2\omega^2=\frac{1}{2}kA^2\]
\[=\frac{1}{2}\times30\times(0.015)^2\]
\[=3.375\times10^{-3}\ \text{J}\approx3.38\times10^{-3}\ \text{J}.\]
(ii) Maximum tension in the spring
The maximum tension occurs at the lowest position, where the extension is greatest:
\[T_{\max}=mg+kA\]
\[=(0.30\times10)+(30\times0.015)\]
\[=3.45\ \text{N}.\]
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