(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]
(a) For a simple pendulum swinging as a simple harmonic oscillator: the maximum potential energy occurs at the two extreme positions of the swing (where the bob is momentarily at rest, farthest from the mean position), and the maximum kinetic energy occurs at the mean (lowest, central) position, where the bob moves fastest.
(b) Comparing \( x = 2.5\sin(0.4\pi t) \) with \( x = A\sin(\omega t) \): amplitude \( A = 2.5\,\text{cm} = 0.025\,\text{m} \), angular frequency \( \omega = 0.4\pi = 1.257\,\text{rad s}^{-1} \).
(i) Maximum velocity: \[ v_{max} = A\omega = 0.025 \times 0.4\pi = 0.025 \times 1.257 = 0.0314\,\text{m s}^{-1}\ (=\pi\,\text{cm s}^{-1}). \]
(ii) Maximum acceleration: \[ a_{max} = A\omega^2 = 0.025 \times (0.4\pi)^2 = 0.025 \times 1.579 = 0.0395\,\text{m s}^{-2}. \]
(c)(i) The quantity \( \sqrt{k/m} \) is the angular frequency \( \omega \) of the oscillation (with \( \omega = 2\pi f \)). It measures how rapidly the mass oscillates: a stiffer spring (larger k) or a smaller mass gives a larger \( \omega \) and hence a higher frequency.
(c)(ii) With \( m = 0.30\,\text{kg}, k = 30\,\text{N m}^{-1}, A = 0.015\,\text{m} \): \( \omega = \sqrt{k/m} = \sqrt{30/0.30} = \sqrt{100} = 10\,\text{rad s}^{-1} \).
Maximum kinetic energy: \[ KE_{max} = \tfrac{1}{2} k A^2 = \tfrac{1}{2} \times 30 \times (0.015)^2 = \tfrac{1}{2} \times 30 \times 2.25 \times 10^{-4} = 3.375 \times 10^{-3}\,\text{J}. \]
Maximum tension in the spring: The static (equilibrium) extension is \( x_0 = mg/k = (0.30 \times 10)/30 = 0.10\,\text{m} \). The greatest tension occurs at the lowest point, where the extension is \( x_0 + A \): \[ T_{max} = k(x_0 + A) = 30 \times (0.10 + 0.015) = 30 \times 0.115 = 3.45\,\text{N}. \]
(a) For a simple pendulum swinging as a simple harmonic oscillator: the maximum potential energy occurs at the two extreme positions of the swing (where the bob is momentarily at rest, farthest from the mean position), and the maximum kinetic energy occurs at the mean (lowest, central) position, where the bob moves fastest.
(b) Comparing \( x = 2.5\sin(0.4\pi t) \) with \( x = A\sin(\omega t) \): amplitude \( A = 2.5\,\text{cm} = 0.025\,\text{m} \), angular frequency \( \omega = 0.4\pi = 1.257\,\text{rad s}^{-1} \).
(i) Maximum velocity: \[ v_{max} = A\omega = 0.025 \times 0.4\pi = 0.025 \times 1.257 = 0.0314\,\text{m s}^{-1}\ (=\pi\,\text{cm s}^{-1}). \]
(ii) Maximum acceleration: \[ a_{max} = A\omega^2 = 0.025 \times (0.4\pi)^2 = 0.025 \times 1.579 = 0.0395\,\text{m s}^{-2}. \]
(c)(i) The quantity \( \sqrt{k/m} \) is the angular frequency \( \omega \) of the oscillation (with \( \omega = 2\pi f \)). It measures how rapidly the mass oscillates: a stiffer spring (larger k) or a smaller mass gives a larger \( \omega \) and hence a higher frequency.
(c)(ii) With \( m = 0.30\,\text{kg}, k = 30\,\text{N m}^{-1}, A = 0.015\,\text{m} \): \( \omega = \sqrt{k/m} = \sqrt{30/0.30} = \sqrt{100} = 10\,\text{rad s}^{-1} \).
Maximum kinetic energy: \[ KE_{max} = \tfrac{1}{2} k A^2 = \tfrac{1}{2} \times 30 \times (0.015)^2 = \tfrac{1}{2} \times 30 \times 2.25 \times 10^{-4} = 3.375 \times 10^{-3}\,\text{J}. \]
Maximum tension in the spring: The static (equilibrium) extension is \( x_0 = mg/k = (0.30 \times 10)/30 = 0.10\,\text{m} \). The greatest tension occurs at the lowest point, where the extension is \( x_0 + A \): \[ T_{max} = k(x_0 + A) = 30 \times (0.10 + 0.015) = 30 \times 0.115 = 3.45\,\text{N}. \]
