Unit 14: Simple Harmonic Motion — Short Questions

Simple harmonic motion (S.H.M.) is a type of oscillatory motion in which the acceleration of a body is always directly proportional to its displacement from the mean position and is always directed towards the mean position. Mathematically, a ∝ −x, or a = −ω²x, where ω is the angular frequency of the motion and the negative sign shows that acceleration and displacement are always oppositely directed.

(i) The restoring force must be directly proportional to the displacement from the mean position. (ii) The restoring force must be proportional to the inertia of the body. (iii) The force and displacement must follow Hooke's law, F = −kx, where k is a constant depending on the system. (iv) The acceleration of the oscillating object must be proportional to the displacement, a ∝ −x, the negative sign showing that acceleration is always directed towards the mean position.

Free oscillations occur when a body oscillates with its own natural frequency without the action of any external force (e.g. a simple pendulum displaced and released); its total energy and amplitude remain constant in the absence of resistive forces. Forced oscillations occur when a freely oscillating system is subjected to an external periodic (driving) force, so the body oscillates at the driving frequency (e.g. a swing pushed repeatedly); here the amplitude depends on the relation between the driving frequency and the natural frequency of the body.

An ideal simple pendulum requires a point mass (bob) of negligible size suspended by a perfectly massless, inextensible string, oscillating with no air resistance or friction. In practice, the bob always has a finite size and mass distribution, the suspending string/wire has some mass and slight elasticity, and air resistance and friction at the support always act on the system. Since these ideal conditions can never be perfectly realised, only an approximation of a simple pendulum (small heavy bob, light string, small amplitude) can be constructed in practice.

The sharpness of resonance depends on the amount of damping present in the oscillating system. The smaller the damping, the greater the amplitude attained at resonance and the sharper (narrower and taller) the resonance peak, since the amplitude rises steeply only very close to the natural frequency. With heavy damping the resonance curve becomes flat and broad, so the resonance is not sharp.

Restoring force is the force that acts on a displaced oscillating body, always directed opposite to its displacement, so as to bring it back towards its mean (equilibrium) position. (i) In a simple pendulum, the component of the bob's weight along the direction of motion, mg sinθ, provides the restoring force. (ii) In a spring, the elastic force of the stretched or compressed spring, F = −kx (Hooke's law), provides the restoring force.

Shock absorbers in a car's suspension system provide critical damping. When the car travels on a bumpy, rough road, the wheels and suspension are repeatedly displaced and tend to oscillate; the shock absorbers quickly dissipate this oscillation energy so that the body of the car returns to its equilibrium position in the shortest possible time, without continuing to bounce up and down, thereby giving the passengers a smooth and comfortable ride.

If soldiers march in step, the frequency of their rhythmic footsteps could become equal to (or an integral multiple of) the natural frequency of the bridge. This would drive the bridge into resonance, causing its amplitude of vibration to grow very large, which could damage or even collapse the bridge. To avoid setting up such resonant vibrations, soldiers are instructed to break step while crossing a bridge.

When the girl stands up, her centre of mass moves upward, closer to the point of suspension of the swing, which effectively reduces the length ℓ of the swinging system (acting like a pendulum). Since T = 2π√(ℓ/g), a decrease in ℓ causes the time period T to decrease, so the swing oscillates with a shorter period (faster) when she stands up.

No, it will not keep accurate time there. The acceleration due to gravity g decreases as altitude increases, so g at Murree, and especially at the top of Mount Everest, is smaller than at Karachi (sea level). Since T = 2π√(ℓ/g), a smaller g gives a larger time period; the pendulum will therefore oscillate more slowly (the clock will run slow / lose time) at Murree, and lose even more time at the top of Mount Everest, compared with Karachi.

Attach a small heavy bob to the free (lower) end of the wire to form a simple pendulum. Displace the bob slightly from its mean position and release it, then use a stopwatch to time a number of complete oscillations and calculate the time period T. Using T = 2π√(ℓ/g), the length is then found from ℓ = gT²/4π². Since the wire's length is very nearly equal to the height of the tower, this gives the length of the tower even though its upper end cannot be seen directly.

The period will increase. For a mass-spring system, T = 2π√(m/k). Since the spring constant k remains unchanged, increasing the attached mass m increases the value of T, i.e. adding more mass makes the oscillations slower (longer period).

No, a simple pendulum cannot vibrate at the centre of the Earth. By symmetry, the gravitational pull of the surrounding mass of the Earth cancels out in every direction at the centre, so the net gravitational acceleration g there is zero. With g = 0, the restoring force on the bob (mg sinθ) is also zero, so there is no restoring force to produce oscillation; also, since T = 2π√(ℓ/g), T would become infinite. Hence the pendulum cannot execute S.H.M. at the centre of the Earth.