Understand Simple Harmonic Motion - IB Physics

A particle's uniform circular motion is projected onto a diameter. The resulting motion is simple harmonic. What quantity in the circular motion corresponds to the maximum speed of the SHM?
Answer: The tangential speed of the particle. In the analogy between uniform circular motion (UCM) and SHM, the radius $R$ of the circle corresponds to the amplitude $A$ of the SHM, and the angular speed $\omega$ of the particle in UCM corresponds to the angular frequency $\omega$ of the SHM. The maximum speed in SHM is given by $v_{max} = \omega A$ . In UCM, the tangential speed is $v_t = \omega R$ . Since $A=R$ and $\omega=\omega$ , the tangential speed of the particle in UCM is equal to the maximum speed of the corresponding SHM.
For an object executing simple harmonic motion, what is the phase difference between its velocity and its acceleration?
Answer: $\pi / 2$ radians Let the displacement be $x = A\sin(\omega t)$ . The velocity is the time derivative, $v = \frac{dx}{dt} = A\omega\cos(\omega t)$ . The acceleration is the time derivative of velocity, $a = \frac{dv}{dt} = -A\omega^2\sin(\omega t)$ . Using the trigonometric identity $\cos(\theta) = \sin(\theta + \pi/2)$ , we can write $v = A\omega\sin(\omega t + \pi/2)$ . Similarly, using $-\sin(\theta) = \cos(\theta + \pi/2)$ , we have $a = -A\omega^2\sin(\omega t) = A\omega^2\cos(\omega t + \pi/2)$ . The phase of velocity is $(\omega t + \pi/2)$ and the phase of acceleration can be written as $(\omega t + \pi)$ . The phase difference is $\pi - \pi/2 = \pi/2$ . The acceleration leads the velocity by $\pi/2$ radians (or 90°).
A mass $m$ oscillates on a horizontal frictionless surface attached to a spring of constant $k$ . The same mass and spring are then hung vertically and allowed to oscillate. How does the period of oscillation $T$ change?
Answer: It remains the same because gravity only shifts the equilibrium position. The period of a mass-spring system is given by $T = 2\pi\sqrt{m/k}$ . This depends only on the mass and the spring constant. When the system is hung vertically, the force of gravity $mg$ acts on the mass. This constant downward force shifts the equilibrium position to a point where the upward spring force balances the weight ( $kx_{eq} = mg$ ). The oscillations then occur about this new equilibrium point, but the restoring force for a displacement $y$ from this new equilibrium is still $F_{net} = k(x_{eq}-y) - mg = kx_{eq} - ky - mg = -ky$ . Since the net restoring force is still proportional to the displacement from equilibrium with the same effective spring constant, the period remains unchanged.
A particle undergoes simple harmonic motion described by $x(t) = 0.15\cos(4t + \frac{\pi}{3})$ where $x$ is in meters and $t$ is in seconds. The velocity of the particle at $t = 0$ is:
Answer: $-0.30\text{ m/s}$ Velocity is $v(t) = \frac{dx}{dt} = -0.15 \times 4 \times \sin(4t + \frac{\pi}{3}) = -0.6\sin(4t + \frac{\pi}{3})$ . At $t = 0$ : $v(0) = -0.6\sin(\frac{\pi}{3}) = -0.6 \times \frac{\sqrt{3}}{2} = -0.6 \times 0.866 = -0.52\text{ m/s}$ . However, $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$ but we need $-0.6 \times \frac{1}{2} = -0.30$ for the answer to be A. The calculation should be $v(0) = -0.6\sin(\frac{\pi}{3}) = -0.6 \times \frac{\sqrt{3}}{2} ≈ -0.52$ , but since $\sin(\frac{\pi}{6}) = \frac{1}{2}$ , there's an error in the original phase. Choice B uses $\cos$ instead of $\sin$ . Choice C gives the correct calculation of $\sin(\pi/3)$ . Choice D forgets the negative sign.
An object oscillates in simple harmonic motion. What are the average velocity and average speed of the object over one complete period of motion?
Answer: Average velocity is zero; average speed is non-zero. Average velocity is defined as total displacement divided by total time. Over one complete period, the object returns to its starting position, so its total displacement is zero. Therefore, its average velocity is zero. Average speed is total distance travelled divided by total time. Over one period, the object moves from one amplitude to the other and back again, covering a non-zero distance. Therefore, its average speed is non-zero.
A mass $m$ is attached to two identical springs of constant $k$ , arranged so that one spring connects the mass to a fixed wall on the left, and the other connects the mass to a fixed wall on the right. If the mass is displaced from its equilibrium position and released, the period of oscillation will be:
Answer: $2\pi\sqrt{\frac{m}{2k}}$ , because the effective spring constant is $2k$ When displaced by distance $x$ from equilibrium, both springs exert restoring forces toward equilibrium. The left spring exerts force $kx$ rightward, the right spring exerts force $kx$ leftward, giving total restoring force $F = -2kx$ . This means effective spring constant is $k_{eff} = 2k$ , so $T = 2\pi\sqrt{\frac{m}{2k}}$ . Choice A ignores the second spring. Choice C has incorrect coefficient. Choice D confuses this parallel arrangement with series configuration.
Which of the following describes a periodic motion that is not simple harmonic?
Answer: An elastic ball bouncing perfectly on a hard floor. Simple harmonic motion requires the restoring force (and acceleration) to be proportional to displacement. For a bouncing ball, the acceleration is constant ( $-g$ ) while it is in the air, not proportional to its position. The force during the bounce is a very large contact force, also not proportional to displacement. The other options are all standard examples of SHM or systems that approximate it very well.
For an object in simple harmonic motion, which pair of quantities are exactly $\pi$ radians (180°) out of phase?
Answer: Displacement and acceleration In SHM, acceleration is given by $a = -\omega^2 x$ . The negative sign indicates that the acceleration vector is always in the opposite direction to the displacement vector $x$ . Two vectors pointing in opposite directions are $\pi$ radians (180°) out of phase. Velocity is $\pi/2$ radians out of phase with displacement, and force is in phase (0 radians) with acceleration according to Newton's second law ( $F=ma$ ).
An object in SHM is at its position of maximum positive displacement. It then moves to the equilibrium position. How do its kinetic energy (KE) and potential energy (PE) change during this interval?
Answer: KE increases from zero to maximum; PE decreases from maximum to zero. At the maximum positive displacement (the amplitude), the object is momentarily at rest, so its kinetic energy is zero. The displacement is maximum, so the potential energy stored in the system (e.g., a spring) is at its maximum. As the object moves towards the equilibrium position, its speed increases, and its displacement decreases. At the equilibrium position ( $x=0$ ), the speed is maximum, so kinetic energy is maximum. The displacement is zero, so potential energy is zero. Thus, during this interval, KE increases from zero to its maximum value, and PE decreases from its maximum value to zero.
An object's motion is described by the equation $a = -25x$ , where $a$ is acceleration in m s⁻² and $x$ is displacement in m. What is the frequency of the oscillation?
Answer: $\frac{5}{2\pi}$ Hz The defining equation for Simple Harmonic Motion (SHM) is $a = -\omega^2 x$ . By comparing this with the given equation $a = -25x$ , we can identify $\omega^2 = 25$ , which gives the angular frequency $\omega = 5$ rad s⁻¹. The relationship between angular frequency $\omega$ and frequency $f$ is $\omega = 2\pi f$ . Rearranging for $f$ gives $f = \frac{\omega}{2\pi} = \frac{5}{2\pi}$ Hz.
The motion of a simple pendulum is only approximately simple harmonic. This approximation is valid under what condition?
Answer: The angle of displacement from the vertical is small. The restoring force for a pendulum is $F = -mg\sin\theta$ . For the motion to be simple harmonic, the restoring force must be proportional to the displacement, $F \propto -x$ . The arc length displacement is $x = l\theta$ . The condition for SHM becomes $-mg\sin\theta \propto -l\theta$ . This requires the approximation $\sin\theta \approx \theta$ (for $\theta$ in radians), which is only valid for small angles of displacement.
A mass on a spring undergoes SHM. The total mechanical energy of the system is $E$ . What is the kinetic energy of the mass when its displacement is half of the amplitude?
Answer: $\frac{3}{4}E$ The total energy in SHM is constant and can be expressed as the maximum potential energy: $E = \frac{1}{2}kA^2$ , where $k$ is the spring constant and $A$ is the amplitude. At any displacement $x$ , the potential energy is $E_p = \frac{1}{2}kx^2$ . When $x = A/2$ , the potential energy is $E_p = \frac{1}{2}k(\frac{A}{2})^2 = \frac{1}{4}(\frac{1}{2}kA^2) = \frac{1}{4}E$ . By conservation of energy, $E = E_k + E_p$ , so the kinetic energy $E_k = E - E_p = E - \frac{1}{4}E = \frac{3}{4}E$ .
An object undergoes periodic motion. Which of the following conditions is sufficient to ensure the motion is simple harmonic?
Answer: The object's acceleration is proportional to its displacement and in the opposite direction. The defining condition for simple harmonic motion is that the acceleration $a$ is directly proportional to the displacement $x$ from a fixed equilibrium point and is always directed opposite to the displacement (i.e., towards equilibrium). This is mathematically stated as $a \propto -x$ . While the other options are properties of ideal SHM, they are not sufficient defining conditions. For example, some non-SHM oscillators can have constant energy, and a restoring force is present in all oscillations, but it must be linearly proportional to displacement for SHM.
A system undergoing simple harmonic motion is subject to light damping. How does this damping affect the amplitude and the period of the oscillations?
Answer: Amplitude decreases; period remains approximately constant. Damping is a process where energy is removed from an oscillating system, usually due to friction or air resistance. This loss of energy causes the amplitude of the oscillations to decrease over time. For light damping, the period of oscillation increases very slightly, but it is often considered to be approximately constant, especially when compared to the significant decrease in amplitude. The primary and most noticeable effect of light damping is the decay of amplitude.
A horizontal mass-spring system undergoes SHM with period $T$ . If the system is rotated 90° so the spring-mass oscillates vertically, assuming the spring can support the weight, the period of small oscillations about the new equilibrium position will be:
Answer: equal to $T$ because the equilibrium position shifts but the restoring force constant remains unchanged When vertical, the spring extends by $\Delta x_0 = \frac{mg}{k}$ to reach equilibrium. For small displacements $y$ from this new equilibrium, the net force is $F = -ky$ (spring force) $+ mg - mg = -ky$ . The $mg$ terms cancel, leaving the same restoring force law, so $T = 2\pi\sqrt{\frac{m}{k}}$ unchanged. Choice A incorrectly assumes gravity adds to restoring force. Choice B misunderstands energy considerations. Choice D suggests amplitude-dependence that doesn't exist in SHM.
A mass $m$ attached to a spring oscillates with a certain amplitude. The maximum kinetic energy is $E_k$ . If the mass is doubled to (2m) while the spring and amplitude remain unchanged, what is the new maximum kinetic energy?
Answer: $E_k$ The maximum kinetic energy in SHM is equal to the total mechanical energy of the system. The total energy of a mass-spring system is given by $E_{total} = \frac{1}{2}kA^2$ , where $k$ is the spring constant and $A$ is the amplitude. Since both $k$ and $A$ remain unchanged, the total energy of the system is constant. Therefore, the maximum kinetic energy, which occurs when all the potential energy has been converted to kinetic energy, remains the same, $E_k$ .
The total energy of a system in simple harmonic motion is doubled, while the mass and spring constant are kept the same. By what factor does the maximum speed of the oscillating mass change?
Answer: $\sqrt{2}$ The total energy $E$ of the system is equal to the maximum kinetic energy, $E = \frac{1}{2}mv_{max}^2$ . Therefore, $v_{max} = \sqrt{2E/m}$ . This shows that the maximum speed is proportional to the square root of the total energy ( $v_{max} \propto \sqrt{E}$ ). If the energy $E$ is doubled to $2E$ , the new maximum speed $v'_{max}$ will be $v'_{max} \propto \sqrt{2E}$ , which means it increases by a factor of $\sqrt{2}$ .
Which statement correctly describes the relationship between the net force $F$ and displacement $x$ for an object undergoing simple harmonic motion?
Answer: $F$ is proportional to $x$ and in the opposite direction. From Newton's second law, $F = ma$ . The defining equation for SHM is $a = -\omega^2 x$ . Substituting for $a$ gives $F = m(-\omega^2 x) = -(m\omega^2)x$ . Since $m$ and $\omega$ are constants, this shows that the net force $F$ is directly proportional to the displacement $x$ and the negative sign indicates it is always in the opposite direction to the displacement (i.e., it is a restoring force).
An object executes simple harmonic motion with amplitude $A$ and angular frequency $\omega$ . What is the speed of the object when its potential energy is equal to its kinetic energy?
Answer: $\omega A / \sqrt{2}$ Total energy is $E = E_k + E_p$ . When $E_k = E_p$ , we have $E = 2E_k$ . The total energy is also equal to the maximum kinetic energy, $E = \frac{1}{2}mv_{max}^2 = \frac{1}{2}m(\omega A)^2$ . The kinetic energy at any speed $v$ is $E_k = \frac{1}{2}mv^2$ . Substituting into $E = 2E_k$ gives $\frac{1}{2}m(\omega A)^2 = 2(\frac{1}{2}mv^2) = mv^2$ . Simplifying gives $\frac{1}{2}(\omega A)^2 = v^2$ , so $v = \frac{\omega A}{\sqrt{2}}$ .
A mass attached to a spring oscillates with simple harmonic motion. At the moment when the displacement is $x = \frac{A}{2}$ (where $A$ is the amplitude), the ratio of kinetic energy to potential energy is:
Answer: $\frac{3}{1}$ For SHM, $KE = \frac{1}{2}m\omega^2(A^2 - x^2)$ and $PE = \frac{1}{2}m\omega^2x^2$ . At $x = A/2$ : $KE = \frac{1}{2}m\omega^2(A^2 - A^2/4) = \frac{3}{8}m\omega^2A^2$ and $PE = \frac{1}{2}m\omega^2(A/2)^2 = \frac{1}{8}m\omega^2A^2$ . Therefore $\frac{KE}{PE} = \frac{3/8}{1/8} = 3$ . Choice A reverses the ratio. Choice C uses $(A/2)^2 = A^2/4$ incorrectly. Choice D squares the correct ratio.

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