Simple Pendulum and Oscillation of a Loaded Spring

Simple Pendulum

A simple pendulum consist of the metal bob of mass ‘m’ which is suspended from a rigid support ‘O’ with the help of the inextensible and weightless string. Let l be the effective length i.e. distance between the point of suspension and center of gravity of the bob of pendulum and A be its mean position. The pendulum A is its mean position. The pendulum of free to oscillate in a vertical plane. The bob is displaced from its mean position by small angle so that the bob will oscillate in a straight line (AB) about the mean position.

The weight (mg) of bob acts vertically downward in the position say B. The height is resolved into two components.

1. Mgcosθ which balances the tension in the string.
2. Mgsinθ which acts towards mean position and is a restoring force.

\begin{align*} F_R &= -mg\sin \theta \\ \text {or,} \: ma &= -mg\sin \theta \end{align*}

Where ‘a’ is acceleration of bob.

$$a = -\sin \theta \dots (i) $$

$$\text {From} \Delta OAB$$

$$\text {here} \sin \theta \approx \theta = \frac yl \dots (ii)$$

Equation (i) becomes

$$ a = -\frac {gy}{l} = -\frac gl y \dots (iii)$$

Comparing Equation (iii) with \(a = -\omega ^2 y;\) we have,

\begin{align*} \omega ^2 &= \frac gl \\ \omega &= \sqrt {\frac gl} \end{align*}

Time period

The time period T of the pendulum is

\begin{align*} T &= \frac {2\pi }{\omega} \\ \therefore T &= 2\pi \sqrt {\frac 1g} \end{align*}

\(a \propto y\) and is directed towards mean position so motion of simple pendulum is simple harmonic.

Hence, time period of a simple pendulum is

1. Independent of the amplitude,
2. Independent of the mass of the bob
3. But depends on the length and the acceleration due to the gravity of a place.

Second's Pendulum

A simple pendulum whose time period is two seconds is called secnd pendulum. i.e. T = 2 sec. On the earth's surface, g = 9.8 m^-2 and from the relation, \(T = 2\pi \sqrt {\frac lg}\).

\begin{align*}2 &= 2\pi \sqrt {\frac {l}{9.8}} \\l &= 0.993 m = 99.3 cm \end{align*}

Hence the length of seconds' pendulum is 99.3 cm.

Vibration of a particle in Horizontal Spring

Let us suppose the spring S with negligible mass which is attached to a wall and the other end to an object of mass, m. The spring S with an object are laid on a horizontal table. If the mass is pulled slightly to extend spring and then released, the system vibrates with simple harmonic motion. The center of oscillation O is the position of mass at the end of the string corresponding to its natural length, i.e. when the spring is neither extended nor compressed.

Let l be the extension of the spring and F be the restoring force set up in the spring. Then from Hook's law

\begin{align*} f &\propto l \\ \text {or.} \: F &=-kl \dots (i) \end{align*}

where k is known as spring constant. Negative sign shows that the restoring force acts opposite to the displacement of the mass.

If a is the acceleration produced in the mass, then we have

$$ F = ma \dots (ii) $$

Therefore, from equation (i) and (ii), we have

\begin{align*} ma &= -kl \\ \text {or,} \: a &= -\frac km l \\ \text {or,} \: a &= -\omega ^2 l\dots (iii) \end{align*}

where \(\omega ^2 = \frac km\) is a constant.

This shows that the acceleration is directly proportional to the displacement and is directed towards the mean position. Hence, the motion of a horizontal mass-spring system is simple harmonic motion.

Expression for time period, T

Now from above equation, we have,

\begin{align*} \omega ^2 = \frac km \\ \text {where} \omega \text {is the angular velocity. If T is the time period of oscillation, then}\\ \omega &= \frac{2\pi }{T}\\ \therefore \omega &= \frac {2\pi }{T} = \sqrt {\frac km} \\ \text {hence,} T &= 2\pi \sqrt {\frac mk} \\ \end{align*}

Which is the required expression for the time period of oscillation depends upon the mass attched to the spring.

Vibration of a particle in Vertical Spring

Consider a light, elastic spring of force constant K. Let its one end be attached to a rigid support such as on ceiling and a mass is attached at the other end as shown in the figure.

Where a load m is attached, the spring extends and let l be the elongation produced as shown in figure. The restoring force on the spring is

$$mg = F_1 = -kl \dots (i) $$

Let the load is pulled down through a small distance y, then the restoring force F₂ is given as

\begin{align*} F_2 &= -K(l +y) \dots (ii) \\ \text {The effective restoring force which causes te oscillation is} \\ F &= F_2 - F_1 \\ &= -K(l + y) -(-kl) \\ &= -ky \\ \text {As} \: F &= ma \\ \therefore ma &= -ky \\ \text {or,} \: a &= -\frac km \times y \\ \text {or,} \: a \propto y \dots (iii) \\ \text {where} \frac km is \text {constant} \end{align*}

Hence, the motion of a loaded vertical spring is simple harmonic.

Time Period

If the load attached on the spring is pulled and lefft, it will start to oscillate. It's time period is calculated as given below:

$$\text {For S.H.M} \: a = -\omega ^2 y $$

For a oscillating loaded spring

\begin{align*} a &= -\frac km y \\ \therefore -\omega ^2 y &= -\frac km y \\ \omega ^2 = \frac km \\ \ \end{align*}where \(\omega \) is the angular velocity. If T is the time period of oscillation, then\begin{align*}\\ \omega &= \frac{2\pi }{T}\\ \therefore \omega &= \frac {2\pi }{T} = \sqrt {\frac km} \\ \text {hence,} T &= 2\pi \sqrt {\frac mk} \\ \end{align*}