Prove that motion of simple pendulum is S.H.M when amplitude is small derive an expression for time period of simple pendulum..
Answers
Explanation:
A simple pendulum has a small-diameter bob and a string that has a very small mass but is strong enough not to stretch appreciably. The linear displacement from equilibrium is s, the length of the arc. Also shown are the forces on the bob, which result in a net force of -\text{mg}\phantom{\rule{0.25em}{0ex}}\text{sin}\theta toward the equilibrium position—that is, a restoring force.
In the figure, a horizontal bar is drawn. A perpendicular dotted line from the middle of the bar, depicting the equilibrium of pendulum, is drawn downward. A string of length L is tied to the bar at the equilibrium point. A circular bob of mass m is tied to the end of the string which is at a distance s from the equilibrium. The string is at an angle of theta with the equilibrium at the bar. A red arrow showing the time T of the oscillation of the mob is shown along the string line toward the bar. An arrow from the bob toward the equilibrium shows its restoring force asm g sine theta. A perpendicular arrow from the bob toward the ground depicts its mass as W equals to mg, and this arrow is at an angle theta with downward direction of string.
Pendulums are in common usage. Some have crucial uses, such as in clocks; some are for fun, such as a child’s swing; and some are just there, such as the sinker on a fishing line. For small displacements, a pendulum is a simple harmonic oscillator. A simple pendulum is defined to have an object that has a small mass, also known as the pendulum bob, which is suspended from a light wire or string, such as shown in [link]. Exploring the simple pendulum a bit further, we can discover the conditions under which it performs simple harmonic motion, and we can derive an interesting expression for its period.
We begin by defining the displacement to be the arc length s. We see from [link] that the net force on the bob is tangent to the arc and equals -\text{mg}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta. (The weight \text{mg} has components \text{mg}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta along the string and \text{mg}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta tangent to the arc.) Tension in the string exactly cancels the component \phantom{\rule{0.25em}{0ex}}\text{mg}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta parallel to the string. This leaves a net restoring force back toward the equilibrium position at \theta =0.
Now, if we can show that the restoring force is directly proportional to the displacement, then we have a simple harmonic oscillator. In trying to determine if we have a simple harmonic oscillator, we should note that for small angles (less than about \text{15º}), \text{sin}\phantom{\rule{0.25em}{0ex}}\theta \approx \phantom{\rule{0.25em}{0ex}}\theta \phantom{\rule{0.25em}{0ex}}(\text{sin}\phantom{\rule{0.25em}{0ex}}\theta and \theta differ by about 1% or less at smaller angles). Thus, for angles less than about \text{15º}, the restoring force F is
F\approx -\text{mg}\mathrm{\theta .}
The displacement s is directly proportional to \theta. When \theta is expressed in radians, the arc length in a circle is related to its radius (L in this instance) by:
s=\mathrm{L\theta },
so that
\theta =\frac{s}{L}.
For small angles, then, the expression for the restoring force is:
F\approx -\frac{\text{mg}}{L}s
This expression is of the form:
F=-\text{kx},
where the force constant is given by k=\text{mg}/L and the displacement is given by x=s. For angles less than about \text{15º}, the restoring force is directly proportional to the displacement, and the simple pendulum is a simple harmonic oscillator.
Using this equation, we can find the period of a pendulum for amplitudes less than about \text{15º}. For the simple pendulum:
T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{m}{\text{mg}/L}}.
Thus,
T=2\pi \sqrt{\frac{L}{g}}