Apply Simple Harmonic Motion - IB Physics

1. An object oscillates with simple harmonic motion of amplitude $x_0$ and angular frequency $\omega$. What is the magnitude of its acceleration when it is at a displacement of $x = x_0/2$?
   Answer: $\omega^2 x_0 / 2$ The defining equation for SHM is $a = -\omega^2 x$. The magnitude of the acceleration is $|a| = \omega^2 |x|$. When the displacement is $x = x_0/2$, the magnitude of the acceleration is $a = \omega^2 (x_0/2) = \omega^2 x_0 / 2$.
2. A particle undergoes simple harmonic motion. When its displacement is at its maximum positive value, what can be said about its velocity and acceleration?
   Answer: Velocity is zero and acceleration is maximum negative. At the maximum positive displacement (the amplitude), the particle momentarily stops before changing direction, so its velocity is zero. According to the defining equation of SHM, $a = -\omega^2x$, acceleration is proportional to the negative of displacement. Therefore, when displacement $x$ is maximum positive, acceleration $a$ is maximum negative.
3. A particle undergoes SHM with angular frequency $\omega = 5.0$ rad/s. At $t = 0$, the particle is at equilibrium moving in the positive direction with speed $v_0 = 2.0$ m/s. What is the phase constant $\phi$ in the equation $x(t) = A\cos(\omega t + \phi)$?
   Answer: $\phi = -\pi/2$ radians, since cosine starts at maximum When analyzing simple harmonic motion problems, you need to match the given initial conditions to the correct form of the position equation. The key is understanding how position, velocity, and phase constants relate at $t = 0$. Given $x(t) = A\cos(\omega t + \phi)$, the velocity is $v(t) = -A\omega\sin(\omega t + \phi)$. At $t = 0$, you have initial conditions: $x(0) = 0$ (at equilibrium) and $v(0) = +2.0$ m/s (positive direction). For the position: $x(0) = A\cos(\phi) = 0$. This means $\cos(\phi) = 0$, which occurs when $\phi = \pm\pi/2$. For the velocity: $v(0) = -A\omega\sin(\phi) = +2.0$ m/s. Since $\omega = 5.0$ rad/s is positive and the velocity is positive, you need $-A\sin(\phi) > 0$. This requires $\sin(\phi) < 0$. When $\phi = -\pi/2$, $\sin(-\pi/2) = -1 < 0$, which gives the correct positive velocity. Therefore, $\phi = -\pi/2$ radians. Option A ($\phi = +\pi/2$) gives $\sin(+\pi/2) = +1$, resulting in negative initial velocity. Option B ($\phi = 0$) gives $\cos(0) = 1$, meaning the particle starts at maximum displacement, not equilibrium. Option D ($\phi = \pi$) gives $\cos(\pi) = -1$, placing the particle at minimum displacement initially. Study tip: Always check both position and velocity initial conditions. The phase constant must satisfy both simultaneously—don't just focus on where the particle starts.
4. An object on a spring is oscillating in a system where light damping is present. Which statement correctly describes the effect of damping on the motion over several oscillations?
   Answer: The amplitude decreases while the period remains approximately constant. Light damping causes a gradual loss of mechanical energy from the oscillating system, resulting in an exponential decrease in the amplitude of oscillation. For light damping, the effect on the period (and thus frequency) is negligible, so it is considered to remain approximately constant. The motion eventually stops, but the frequency doesn't decrease to zero during the oscillation.
5. A student wants to construct a simple pendulum with a period of 1.0 s. Approximately what length should the pendulum have?
   Answer: 0.25 m The formula for the period of a simple pendulum is $T = 2\pi\sqrt{l/g}$. We can rearrange this to solve for the length $l$: $T^2 = (2\pi)^2(l/g)$, so $l = gT^2 / (2\pi)^2$. Using $g \approx 9.8 \text{ m s}^{-2}$ and $T = 1.0 \text{ s}$, we get $l = (9.8 \cdot 1.0^2) / (4\pi^2) \approx 9.8 / (4 \cdot 9.87) \approx 9.8 / 39.5 \approx 0.248 \text{ m}$. This is approximately 0.25 m.
6. A mass attached to a vertical spring oscillates with SHM. The spring constant is $k = 200$ N/m and the mass is $m = 0.50$ kg. If the amplitude of oscillation about the equilibrium position is $A = 0.075$ m, what is the maximum acceleration experienced by the mass?
   Answer: $30$ m/s² occurring when the spring force is maximum In SHM, acceleration a = -ω²x where ω = √(k/m) = √(200/0.50) = √400 = 20 rad/s. Maximum acceleration occurs at maximum displacement: |a|_max = ω²A = (20)² × 0.075 = 400 × 0.075 = 30 m/s². The maximum acceleration occurs at maximum displacement (turning points) when the spring force is maximum. Choice A incorrectly states maximum acceleration occurs at equilibrium (where a = 0). Choice B has the wrong magnitude. Choice C has wrong magnitude but mentions the correct location.
7. The total energy of a mass-spring system in simple harmonic motion is doubled. By what factor does the amplitude of the oscillation change?
   Answer: $\sqrt{2}$ The total energy $E$ of a mass-spring system in SHM is proportional to the square of the amplitude $A$, given by $E = \frac{1}{2}kA^2$. If the energy is doubled to $2E$, then $2E = \frac{1}{2}kA'^2$. Taking the ratio of the new energy to the old energy gives $2 = (A'/A)^2$. Therefore, the new amplitude $A'$ is $A\sqrt{2}$. The factor of change is $\sqrt{2}$.
8. A mass attached to a vertical spring is at its lowest point in its oscillation. Which statement is correct about the net force and acceleration at this instant?
   Answer: Both the net force and acceleration are at their maximum upward value. At the lowest point of oscillation, the displacement from the equilibrium position is maximum downwards. According to the SHM definition $a = -\omega^2 x$, if $x$ is maximum negative, acceleration $a$ is maximum positive (upwards). By Newton's second law $F_{net} = ma$, the net force must also be maximum and in the upward direction. This is because the upward spring force is at its maximum and is greater than the downward force of gravity.
9. Two pendulums A and B have the same length but different masses ($m_A = 2m_B$). Both are displaced by the same small angle and released simultaneously. After pendulum A completes exactly 10 oscillations, how many oscillations has pendulum B completed?
   Answer: Exactly 10 oscillations, since period is independent of mass For a simple pendulum, the period T = 2π√(L/g) depends only on length L and gravitational acceleration g, not on mass m. Since both pendulums have identical length, they have identical periods and frequencies. Therefore, they complete oscillations at exactly the same rate. After any given time interval, both will have completed the same number of oscillations. Choice B incorrectly assumes mass affects period. Choice C shows the common misconception that lighter objects oscillate faster. Choice D incorrectly suggests amplitude dependence for small oscillations.
10. The velocity $v$ of an object in SHM is related to its displacement $x$ by $v = \pm \omega \sqrt{x_0^2 - x^2}$. At which displacement is the object's speed equal to half its maximum speed?
    Answer: $x_0 \sqrt{3} / 2$ The maximum speed $v_{max}$ occurs at $x=0$, so $v_{max} = \omega x_0$. We want to find the displacement $x$ where the speed $v = v_{max}/2 = \omega x_0 / 2$. Using the given equation: $\omega x_0 / 2 = \omega \sqrt{x_0^2 - x^2}$. Dividing by $\omega$ gives $x_0 / 2 = \sqrt{x_0^2 - x^2}$. Squaring both sides: $x_0^2 / 4 = x_0^2 - x^2$. Rearranging for $x^2$: $x^2 = x_0^2 - x_0^2 / 4 = 3x_0^2 / 4$. Taking the square root gives $x = x_0 \sqrt{3} / 2$.
11. A simple pendulum is suspended from the ceiling of an elevator. The elevator is accelerating upwards at a constant rate. How does the period of the pendulum's oscillation compare to its period when the elevator is at rest?
    Answer: The period is less. When the elevator accelerates upwards, the apparent weight of the pendulum bob increases. This is equivalent to an increase in the effective gravitational field strength, $g_{eff} = g + a$. Since the period of a pendulum is $T = 2\pi\sqrt{l/g_{eff}}$, an increase in $g_{eff}$ results in a decrease in the period $T$.
12. A mass $m$ attached to a spring with spring constant $k$ oscillates with period $T$. What is the period of oscillation if the mass is changed to (2m) and the spring constant is changed to $k/2$?
    Answer: $2T$ The period of a mass-spring system is $T = 2\pi\sqrt{m/k}$. The new mass is $m' = 2m$ and the new spring constant is $k' = k/2$. The new period is $T' = 2\pi\sqrt{m'/k'} = 2\pi\sqrt{2m/(k/2)} = 2\pi\sqrt{4m/k} = 2 \cdot (2\pi\sqrt{m/k}) = 2T$.
13. A pendulum oscillates with simple harmonic motion. When the pendulum bob is at 60% of its maximum displacement from equilibrium, what is the ratio of its kinetic energy to its total mechanical energy?
    Answer: 0.64 In SHM, total energy E = ½kA² where A is amplitude. At displacement x = 0.6A, potential energy U = ½kx² = ½k(0.6A)² = 0.36(½kA²) = 0.36E. Since energy is conserved, kinetic energy K = E - U = E - 0.36E = 0.64E. Therefore K/E = 0.64. Choice A gives the potential energy ratio. Choice C incorrectly uses the displacement ratio directly. Choice D uses an incorrect energy relationship.
14. An object in simple harmonic motion has a period of 4.0 s. What is the minimum time it takes for the object to travel from its maximum displacement to the equilibrium position?
    Answer: 1.0 s One full period (T) is the time taken to complete one full oscillation. This can be divided into four equal parts: from maximum displacement to equilibrium, from equilibrium to minimum displacement, from minimum displacement back to equilibrium, and from equilibrium back to maximum displacement. Each of these segments takes a time of T/4. Therefore, the time taken is $4.0 \text{ s} / 4 = 1.0 \text{ s}$.
15. The motion of a simple pendulum is only approximately simple harmonic. Under which condition does its motion most closely resemble SHM?
    Answer: When the amplitude of the swing is very small. The restoring force for a pendulum is $F = -mg\sin\theta$. Simple harmonic motion requires the restoring force to be proportional to displacement, $F \propto -x$ (or $F \propto -\theta$ for angular motion). The small-angle approximation states that for small angles $\theta$, $\sin\theta \approx \theta$. Therefore, the pendulum's motion is approximately SHM only when the amplitude of the swing is small.
16. A mass-spring system has total energy $E = 0.50$ J and amplitude $A = 0.10$ m. At what position will the particle's speed equal 75% of its maximum speed?
    Answer: $\pm 0.066$ m from equilibrium Maximum speed occurs at equilibrium: v_max = ωA. When v = 0.75v_max, we use energy conservation: ½mv² + ½kx² = ½mω²A². Substituting v = 0.75ωA: ½m(0.75ωA)² + ½kx² = ½mω²A². This gives ½m(0.5625ω²A²) + ½kx² = ½mω²A². Since k = mω², we get: 0.5625(½mω²A²) + ½mω²x² = ½mω²A². Dividing by ½mω²: 0.5625A² + x² = A². Therefore x² = A²(1 - 0.5625) = 0.4375A². So |x| = A√0.4375 = 0.10√0.4375 ≈ 0.066 m. Choice B uses √0.5. Choice C uses an incorrect energy fraction. Choice D uses 0.75² incorrectly.
17. An object of mass 0.50 kg undergoes SHM with a frequency of 2.0 Hz and an amplitude of 0.10 m. What is the magnitude of the maximum force on the object?
    Answer: 7.9 N First, calculate the angular frequency: $\omega = 2\pi f = 2\pi(2.0 \text{ Hz}) = 4\pi \text{ rad s}^{-1}$. Next, find the maximum acceleration: $a_{max} = \omega^2 A = (4\pi)^2(0.10 \text{ m}) = 1.6\pi^2 \text{ m s}^{-2}$. Finally, use Newton's second law to find the maximum force: $F_{max} = m a_{max} = (0.50 \text{ kg})(1.6\pi^2 \text{ m s}^{-2}) = 0.8\pi^2 \text{ N} \approx 7.9 \text{ N}$.
18. A simple harmonic oscillator has maximum kinetic energy $K_{max}$ and maximum potential energy $U_{max}$. What is the kinetic energy of the oscillator when its displacement is half of its amplitude?
    Answer: $3K_{max} / 4$ The total energy is $E_T = K_{max} = U_{max} = \frac{1}{2}kA^2$. The potential energy at displacement $x$ is $U = \frac{1}{2}kx^2$. At $x = A/2$, the potential energy is $U = \frac{1}{2}k(A/2)^2 = \frac{1}{4}(\frac{1}{2}kA^2) = U_{max}/4$. By conservation of energy, $K + U = E_T$. Therefore, $K = E_T - U = U_{max} - U_{max}/4 = 3U_{max}/4$. Since $U_{max} = K_{max}$, the kinetic energy is $3K_{max}/4$.
19. Which statement correctly describes the relationship between the acceleration $a$ and displacement $x$ of an object in simple harmonic motion?
    Answer: Acceleration is proportional to displacement and is in the opposite direction. The defining equation of simple harmonic motion is $a = -\omega^2x$. This equation shows that acceleration $a$ is directly proportional to the displacement $x$ (since $\omega^2$ is a positive constant). The negative sign indicates that the acceleration vector always points in the opposite direction to the displacement vector.
20. A mass on a spring undergoes SHM with amplitude $A = 0.12$ m and frequency $f = 2.0$ Hz. At what distance from equilibrium will the acceleration magnitude equal half the maximum acceleration?
    Answer: 0.060 m from equilibrium position In SHM, acceleration a = -ω²x where ω = 2πf. Maximum acceleration occurs at maximum displacement: a_max = ω²A. When |a| = ½a_max, we have ω²|x| = ½ω²A, so |x| = A/2 = 0.12/2 = 0.060 m. Choice A uses ¼ of amplitude incorrectly. Choice C uses A/√2 which applies to energy relationships. Choice D uses an incorrect trigonometric relationship.

Start this quiz