Basic Concepts of Simple Harmonic Motion

# Article: Basic Concepts of Simple Harmonic Motion

Simple Harmonic Motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement from its equilibrium position and acts in the direction opposite to the displacement. SHM can be observed in various physical systems, such as mass-spring systems, pendulums under small angle approximation, and even in the motion of celestial bodies under certain conditions.

## Characteristics of SHM

### 1. Restoring Force
The most crucial aspect of SHM is the restoring force. It always points towards the equilibrium position and has a magnitude that is proportional to the displacement from equilibrium. Mathematically, this is often represented by Hooke’s Law, which in its simplest form is:
\[ F = -kx \]
\( F \) is the restoring force,
\( k \) is the spring constant or force constant, and
\( x \) is the displacement from the equilibrium position.

### 2. Periodicity
SHM is periodic, which means it repeats itself after a fixed time interval called the period (T). The period of SHM depends on the properties of the system, such as mass and spring constant, but not on the amplitude of oscillation.

### 3. Amplitude
The amplitude (A) is the maximum displacement from the equilibrium position. It is one of the factors that determine the energy stored in the oscillating system.

### 4. Frequency and Angular Frequency
The frequency (f) of the SHM is the number of oscillations per unit time. The angular frequency (ω) is related to the frequency by the equation:
\[ \omega = 2\pi f \]
The angular frequency of SHM is given by:
\[ \omega = \sqrt{\frac{k}{m}} \]
\( m \) is the mass of the object.

### 5. Energy in SHM
The total mechanical energy in SHM is conserved and is the sum of potential and kinetic energies. The potential energy is maximum at the extremities of the motion, while the kinetic energy is maximum at the equilibrium position.

### 6. Phase and Phase Constant
The phase of SHM indicates the state of the oscillation. It is given by:
\[ \phi = \omega t + \phi_0 \]
\( \phi_0 \) is the initial phase at time \( t = 0 \).

## Equations of Motion for SHM
The displacement \( x \) as a function of time \( t \) in SHM can be described as:
\[ x(t) = A \cos(\omega t + \phi_0) \]
The velocity \( v \) and acceleration \( a \) can be derived respectively as:
\[ v(t) = -A\omega \sin(\omega t + \phi_0) \]
\[ a(t) = -A\omega^2 \cos(\omega t + \phi_0) \]
Notice that acceleration is also proportional to the displacement but in the opposite direction, which maintains the SHM.

# Problems and Solutions on Simple Harmonic Motion

1. **Problem:** Calculate the period of a 0.5 kg mass attached to a spring with a spring constant of 200 N/m.

**Solution:** Use the formula for angular frequency \( \omega \) and the relation between period \( T \) and angular frequency \( \omega \).

\[ \omega = \sqrt{\frac{k}{m}}=\sqrt{\frac{200}{0.5}}=20\,\text{rad/s} \]
\[ T = \frac{2\pi}{\omega} = \frac{2\pi}{20} = \frac{\pi}{10}\,\text{s} \]

2. **Problem:** A mass oscillates with an amplitude of 0.1 m. Calculate its maximum velocity if its period is 2 s.

**Solution:** The maximum velocity \( v_{max} \) occurs at the equilibrium position and is given by:

\[ v_{max} = A\omega = A\left(\frac{2\pi}{T}\right) = 0.1 \left(\frac{2\pi}{2}\right) = 0.1\pi\,\text{m/s} \]

3. **Problem:** Find the maximum acceleration of an object in SHM with a frequency of 0.5 Hz and an amplitude of 0.2 m.


\[ a_{max} = A\omega^2 = A (2\pi f)^2 = 0.2 (2\pi \cdot 0.5)^2 = 0.2\pi^2\,\text{m/s}^2 \]

4. **Problem:** If a block attached to a spring has an elastic potential energy of 1 J at a displacement of 0.1 m from equilibrium, calculate the spring constant.

**Solution:** Use the formula for potential energy in SHM.

\[ U = \frac{1}{2}kx^2 \Rightarrow k = \frac{2U}{x^2} = \frac{2\cdot 1}{0.1^2} = 200\,\text{N/m} \]

5. **Problem:** Calculate the period of a pendulum that is undergoing SHM with a length of 0.5 m. (Assume the acceleration due to gravity, \( g = 9.81\,\text{m/s}^2 \).)

**Solution:** For a simple pendulum, the period \( T \) is given by:

\[ T = 2\pi\sqrt{\frac{L}{g}} = 2\pi\sqrt{\frac{0.5}{9.81}} \approx 1.42\,\text{s} \]

6. **Problem:** An object in SHM has a displacement of 0.1 m at \( t = 1 \) s. If its angular frequency is 5 rad/s, what is its initial phase constant \( \phi_0 \) assuming the amplitude is 0.2 m?

**Solution:** Use the displacement equation for SHM.

\[ x(t) = A\cos(\omega t + \phi_0) \Rightarrow \phi_0 = \cos^{-1}\left(\frac{x}{A}\right) – \omega t \]
\[ \phi_0 = \cos^{-1}\left(\frac{0.1}{0.2}\right) – 5\cdot 1 \approx -4.67\,\text{rad} \]

7. **Problem:** What is the displacement of a 0.3 kg mass in SHM with a spring constant of 180 N/m at \( t = 2 \) s, assuming the initial phase \( \phi_0 \) is zero and the amplitude is 0.1 m?

**Solution:** Determine the angular frequency and use the displacement equation.

\[ \omega = \sqrt{\frac{k}{m}}=\sqrt{\frac{180}{0.3}}=24.49\,\text{rad/s} \]
\[ x(t) = A\cos(\omega t) = 0.1\cos(24.49\cdot 2) \approx -0.1\,\text{m} \]

8. **Problem:** If a mass-spring system has a kinetic energy of 0.5 J at its equilibrium position, with a mass of 0.2 kg, calculate the angular frequency.

**Solution:** The kinetic energy at the equilibrium position equals the total energy.

\[ E = \frac{1}{2}mv^2 = \frac{1}{2}m(A\omega)^2 \Rightarrow \omega = \sqrt{\frac{2E}{mA^2}} \]
Since we don’t have \( A \), further information would be needed to solve this problem.

9. **Problem:** A pendulum clock keeps time accurately on Earth (with \( g = 9.81\,\text{m/s}^2 \)). If taken to the moon where \( g_{moon} = 1.62\,\text{m/s}^2 \), by what factor will its period change?

**Solution:** The period on Earth \( T_{Earth} \) and on the moon \( T_{Moon} \) is proportional to \( \sqrt{g} \).

\[ T_{Moon} = T_{Earth}\sqrt{\frac{g_{Earth}}{g_{Moon}}} \]
\[ \text{Factor} = \sqrt{\frac{9.81}{1.62}} \approx 2.46 \]
The period on the moon would be approximately 2.46 times longer than on Earth.

10. **Problem:** Calculate the displacement from the equilibrium position after half a period has passed for an SHM system with an amplitude of 0.4 m.

**Solution:** After half a period, the object is at the opposite extremity.

\[ x(T/2) = -A = -0.4\,\text{m} \]

11. **Problem:** If a particle is undergoing SHM and the maximum speed is \( 0.5\,\text{m/s} \) with a frequency of \( 1\,\text{Hz} \), find the amplitude of oscillation.

**Solution:** The formula for the maximum speed in SHM is \( v_{max} = A\omega \).

\[ A = \frac{v_{max}}{\omega} = \frac{0.5}{2\pi\cdot 1} \approx 0.0796\,\text{m} \]

12. **Problem:** A mass attached to a spring oscillates with a period of 3 s and an amplitude of 5 cm. Determine the spring constant if the mass is 250 g.

**Solution:** Convert the mass to kilograms and use the formula \( T = 2\pi\sqrt{\frac{m}{k}} \).

\[ k = \frac{4\pi^2 m}{T^2} = \frac{4\pi^2 \cdot 0.25}{3^2} \approx 0.876\,\text{N/m} \]

13. **Problem:** What is the displacement of an SHM oscillator, 4 seconds after it starts from its equilibrium position with a speed of 0.3 m/s and a spring constant of 12 N/m?

**Solution:** First, calculate the mass using \( v = A\omega \) where \( A \) is the amplitude and \( m \) is the mass.

\[ \omega = \sqrt{\frac{k}{m}} \Rightarrow m = \frac{k}{\omega^2} = \frac{k}{\left(\frac{v}{A}\right)^2} \]

Since we don’t have the amplitude \( A \) or the angular frequency \( \omega \), additional information is needed to solve this.

14. **Problem:** How long does it take for a mass-spring system with a mass of 0.150 kg and a spring constant of 100 N/m to move from its equilibrium position to \( \frac{1}{2} \) of its amplitude of 0.2 m?

**Solution:** First find the angular frequency and then the time for that particular displacement.

\[ \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{100}{0.150}} \approx 25.82\,\text{rad/s} \]
\[ x(t) = A\cos(\omega t) \Rightarrow t = \frac{1}{\omega}\cos^{-1}\left(\frac{x}{A}\right)
\[ t = \frac{1}{\omega}\cos^{-1}\left(\frac{0.1}{0.2}\right) = \frac{1}{25.82}\cos^{-1}(0.5) \approx 0.0605\,\text{s} \]

15. **Problem:** If the energy of a 0.1 kg mass undergoing SHM is 0.004 J, what is its speed when the displacement is 3 cm given the spring constant is 80 N/m?

**Solution:** Use the formula for the total energy and solve for speed.

\[ E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 \Rightarrow v = \sqrt{\frac{2E}{m} – \frac{kx^2}{m}}
\[ v = \sqrt{\frac{2\cdot 0.004}{0.1} – \frac{80\cdot 0.03^2}{0.1}} \approx 0.28\,\text{m/s} \]

16. **Problem:** Find the maximum acceleration of a 0.5 kg mass undergoing SHM when attached to a spring with a constant of 50 N/m.

**Solution:** Use the equation \( a_{max} = A\omega^2 \) where \( A \) is the amplitude.

Since no amplitude is given, we can only provide an equation based on the angular frequency.

\[ a_{max} = A\left(\sqrt{\frac{k}{m}}\right)^2 = A\cdot\frac{k}{m} \]

Further information is required to proceed with a numerical answer.

17. **Problem:** A pendulum with a 25 cm long string has a period of 1 s. What is the gravitational acceleration?

**Solution:** Use the formula for the period of a pendulum.

\[ T = 2\pi\sqrt{\frac{L}{g}} \Rightarrow g = \frac{(2\pi)^2 L}{T^2}
\[ g = \frac{(2\pi)^2 \cdot 0.25}{1^2} \approx 9.87\,\text{m/s}^2 \]

18. **Problem:** A block attached to a spring with a spring constant of 150 N/m has a velocity of 1.5 m/s at a displacement of 0.1 m from the equilibrium. Calculate the mass of the block.

**Solution:** Use the conservation of energy principle, where the total energy is the sum of kinetic and potential energies.

\[ E = \frac{1}{2}kx^2 + \frac{1}{2}mv^2 \Rightarrow m = \frac{2E – kx^2}{v^2}
To solve this problem, the energy \( E \) of the system needs to be known or it can be calculated based on the information provided at a specific position (max displacement for total potential energy, or equilibrium for total kinetic energy).

19. **Problem:** An oscillator has a maximum speed of 20 cm/s and a maximum acceleration of 4 m/s^2. Find its period \( T \).

**Solution:** The maximum acceleration and speed relate to amplitude and angular frequency by \( a_{max} = A\omega^2 \) and \( v_{max} = A\omega \).

\[ \frac{a_{max}}{v_{max}} = \omega \Rightarrow \omega = \frac{4}{0.2} = 20\,\text{s}^{-1}
\[ T = \frac{2\pi}{\omega} = \frac{2\pi}{20} = \frac{\pi}{10}\,\text{s} \]

20. **Problem:** Determine the kinetic energy of a 0.05 kg particle in SHM at a position where its displacement is one-quarter of the amplitude if the amplitude is 0.08 m and the spring constant is 200 N/m.

**Solution:** First find the angular frequency and then the kinetic energy.

\[ \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{0.05}} = 89.44\,\text{rad/s} \]
\[ KE = \frac{1}{2}mv^2 = \frac{1}{2}m\left(A\omega \sin(\omega t)\right)^2 \]
\[ \sin(\omega t) = \sqrt{1 – \cos^2(\omega t)} = \sqrt{1 – \left(\frac{1}{4}\right)^2} = \sqrt{1 – \frac{1}{16}} = \sqrt{\frac{15}{16}} \]
\[ KE = \frac{1}{2}\cdot 0.05 \left(0.08\cdot 89.44 \cdot \sqrt{\frac{15}{16}}\right)^2 \approx 0.199\,\text{J} \]

These problems illustrate the application of formulas and concepts related to SHM. To solve them correctly, understanding the underlying physics and careful manipulation of the equations are both required.

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