What are Some Examples of Simple Harmonic Motion?

The study of Physics Topics involves the exploration of matter, energy, and the forces that govern the universe.

Is The Period of a Vertical Spring Same as Horizontal Spring?

1. Oscllation of a mass attached to a vertical elastic spring: Let a body of mass rn be attached to the bottom of a vertical elastic spring of negligible mass suspended from a rigid support [Fig.]. As a result of this, let the increase in length of the spring be l.

So the force constant of the spring,
k = force required for a unit increase in length = \(\frac{m g}{l}\)

Now the mass is pulled downwards through a distance x from its position of equilibrium O. If the extension of the spring does not exceeds its elastic limit, then a reaction force, -kx, equal and opposite to the applied force is developed in the spring. This force acts as the restoring force. If a is the acceleration of the suspended body, then restoring force
-kx = ma
or, a = \(\frac{-k}{m}\)x = -ω²x [where ω = \(\sqrt{\frac{k}{m}}\)]

As the motion of the body of mass m obeys the equation a = -ω²x, it can be said that the motion of the body attached to the spring is simple harmonic.

In this case, time period of oscillation,
T = \(\frac{2 \pi}{\omega}\) = 2π\(\sqrt{\frac{m}{k}}\)
Now, k = \(\frac{m g}{l}\); therefore T = 2π\(\sqrt{\frac{m}{\frac{m g}{l}}}\) = 2π\(\sqrt{\frac{l}{g}}\)

Here, the initial increase in length of the spring due to suspension of the body of mass m is l. So by measuring this increase in length with a metre scale and the time period T with a stopwatch, acceleration due to gravity g can be calculated from the above relation.

2. Oscillation of a mass attached to a horizontal elastic spring: Let one end of an elastic spring of negligible mass be attached to a vertical support and its other end to a body of mass m [Fig.(a)]. The body lies on a smooth horizontal plane. At this moment, no force acts on the body due to the spring as it is not stretched. So the body is at rest.

If the body is now moved towards the right, the spring will be elongated and a restoring force F will act on the body towards the left [Fig.(b)], trying to bring the mass to its equilibrium position.

If the force constant of the spring is k and the body is moved through a distance x towards right, then F = – kx.

∴ Acceleration of the body,
a = \(\frac{F}{m}\) = \(\frac{-k x}{m}\) = -ω²x [where ω = \(\sqrt{\frac{k}{m}}\)]
As the motion of the body obeys the equation, a = -ω²x, it can be said that the motion of the body attached to the spring is simple harmonic.
Time period of oscillation, T = \(\frac{2 \pi}{\omega}\) = 2π\(\sqrt{\frac{m}{k}}\)

It is to be noted that the time periods of vertical oscillation and horizontal oscillation of a spring are equal.

3. Oscillation of a liquid in a U-tube: Consider that a U-tube of uniform cross-section α [Fig] contains a liquid of density ρ. Let the length of the liquid column in each limb at equilibrium be L. Therefore the total length of the liquid column is 2L, if the horizontal separation between the two limbs is negligibly small.

Then the mass of the Liquid column, m = 2Lαρ
If the liquid in one limb is depressed by x then the liquid level in the other limb will be raised by x. Hence the difference in the height of the liquid levels in the two limbs will be 2x.

Weight of this liquid head 2xαρg. This weight provides a restoring force trying to bring the liquid to its initial equilibrium. This force acts opposite to the direction of displacement x in the two limbs.
Thus, restoring force = -2xαρg.
Due to this force, if a is the acceleration of the liquid level, then
ma = -2xαρg
or, 2Lαρa = -2xαρg
or, a = \(\frac{-g \cdot x}{L}\) = -ω²x [where ω = \(\sqrt{\frac{g}{L}}\)]

As the motion of the liquid level obeys the equation a = -ω²x, it follows a simple harmonic motion. So, if the liquid in one limb of a U-tube is depressed and then released, the up and down motion of the liquid column would be simple harmonic.

Time period of this motion, T = \(\frac{2 \pi}{\omega}\) = 2π\(\sqrt{\frac{L}{g}}\)

4. Oscillation of a piston in a gas cylinder: Suppose, some amount of gas is enclosed in a cylinder fitted with a frictionless piston [Fig.]. Let the piston be initially at C, the position of equilibrium, the pressure of the gas enclosed be P and the length of the gas column be L. The piston is now pushed down slightly to B very slowly and then released. The compressed gas will then expand and cause the piston to oscillate up and down.

When the piston is moved through a distance x-from C to B, suppose the pressure of the enclosed gas increases from P to P+ p and the volume decreases to (V – v) from V. If this change takes place isothermally, then according to Boyle’s law,
PV = (P + p) (V – v)
or, PV = PV – Pv + pv – pv
or, Pv = pV [neglecting pv as it is very small]
or, p = \(\frac{P v}{V}\) = \(\frac{P \alpha x}{\alpha L}\) = \(\frac{P x}{L}\);
[α = cross-sectional area of the piston]

An additional force acts on the piston for this excess pressure p and tries to bring the piston to its initial position of equilibrium.

So, the restoring force = -pα = \(\frac{-P \alpha x}{L}\)
a \(=\frac{\text { restoring force }}{\text { mass of the piston }}\) = \(\frac{-P \alpha x}{LM}\)
[M = mass of the piston]
or, a = -ω²x, where ω = \(\sqrt{\frac{P \alpha}{L M}}\)
The motion of the piston obeys the equation a = -ω²x. So this motion is simple harmonic.
Time period of this motion,
T = \(\frac{2 \pi}{\omega}\) = 2π\(\sqrt{\frac{L M}{P \alpha}}\) = 2π\(\sqrt{\frac{L \alpha M}{P \alpha^2}}\) = 2π\(\sqrt{\frac{V M}{P \alpha^2}}\)