## 17.2 Simple Harmonic Motion Simple Harmonic Motion (SHM) is a specific type of periodic motion where a system oscillates about an equilibrium (mean) position. It is defined by a restoring force that is directly proportional to the displacement from this equilibrium position. ### Definition of SHM Simple Harmonic Motion is an oscillatory motion where the acceleration of a particle is directly proportional to its displacement from the mean position and is always directed towards the mean position. - Mathematically, this is expressed as: $$a \propto -x$$ The negative sign indicates that the acceleration vector always points in the opposite direction to the displacement vector. ### Example: Mass on a Spring A classic example of SHM is a mass *m* attached to an elastic spring on a frictionless horizontal surface. - **Equilibrium Position (x = 0):** When the spring is neither stretched nor compressed, the mass is at its mean or equilibrium position (Fig. 17.3 b). - **Restoring Force:** If the mass is displaced by *x* (e.g., stretched to the right, Fig. 17.3 a), the spring exerts a restoring force to bring it back to equilibrium. This force follows **Hooke's Law**: $$F = -kx$$ - *F* is the restoring force. - *k* is the spring constant, a measure of the spring's stiffness (SI unit: $\mathrm{N m^{-1}}$). - *x* is the displacement from the equilibrium position. - The negative sign signifies that the force is always directed opposite to the displacement. When the displaced mass is released, the restoring force accelerates it toward the equilibrium position. Due to inertia, the mass overshoots the equilibrium position and compresses the spring (Fig. 17.3 c). The compressed spring then pushes it back, and the process repeats, causing the mass to oscillate back and forth. ### Dynamics of SHM By combining Newton's Second Law of Motion ($F = ma$) with Hooke's Law ($F = -kx$), we can describe the motion: $$ma = -kx$$ Rearranging for acceleration gives the fundamental equation for SHM in a mass-spring system: $$a = -\frac{k}{m}x$$ This equation confirms that acceleration *a* is directly proportional to displacement *x* and is oppositely directed, satisfying the definition of SHM. ### Frequency and Time Period For a mass-spring system, the term $\frac{k}{m}$ is a constant related to the angular frequency ($\omega$) of the oscillation. | Quantity | Formula | Description | | :--- | :--- | :--- | | **Angular Frequency ($\omega$)** | $\omega = \sqrt{\frac{k}{m}}$ | The rate of change of the phase of the sinusoid, measured in rad/s. | | **Time Period (T)** | $T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m}{k}}$ | The time taken for one complete oscillation. | | **Frequency (f)** | $f = \frac{1}{T} = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$ | The number of complete oscillations per second. | ## Worked Examples **Example 1:** A spring stretches 8.0 cm for a 10.0 kg load. What is the spring constant of the spring? **Solution:** First, convert displacement to meters: $x = 8.0 \text{ cm} = 0.08 \text{ m}$. The force applied is the weight of the load: $F = W = mg = 10.0 \text{ kg} \times 9.8 \text{ m/s}^2 = 98 \text{ N}$. Using Hooke's Law ($F = kx$), the spring constant is: $$k = \frac{F}{x} = \frac{98}{0.08} = 1225 \text{ N m}^{-1}$$ **Example 2:** What is the mass of a fish that stretches the same spring ($k = 1225 \text{ N m}^{-1}$) by 5.5 cm? **Solution:** Convert displacement to meters: $x = 5.5 \text{ cm} = 0.055 \text{ m}$. The force is $F = mg$ and also $F = kx$. Therefore, $mg = kx$. Solving for mass: $$m = \frac{kx}{g} = \frac{1225 \times 0.055}{9.8} \approx 6.9 \text{ kg}$$ **Example 3:** An object with mass 500 g is suspended from a spring, which is stretched by 9.8 cm. Calculate the spring constant. **Solution:** Mass $m = 500 \text{ g} = 0.5 \text{ kg}$. Displacement $x = 9.8 \text{ cm} = 0.098 \text{ m}$. The force is the object's weight: $F = mg = 0.5 \times 9.8 = 4.9 \text{ N}$. The spring constant is: $$k = \frac{F}{x} = \frac{4.9}{0.098} = 50 \text{ N m}^{-1}$$