Advertisements
Advertisements
प्रश्न
A hollow sphere of radius 2 cm is attached to an 18 cm long thread to make a pendulum. Find the time period of oscillation of this pendulum. How does it differ from the time period calculated using the formula for a simple pendulum?
Advertisements
उत्तर
\[\frac{18}{100} = 0 . 18 m = 0 . 2 m\]
Let I be the moment of inertia and \[\omega\] be the angular speed.
Using the energy equation, we can write:
\[mgl(1 - \cos \theta) + \frac{1}{2}I \omega^2 = \text { constant }\]
\[mg\left( 0 . 20 \right) \left( 1 - \cos \theta \right) + \frac{1}{2}I \omega^2 = C . . . \left( 1 \right)\]
\[\text { Moment of inertia about the point of suspension A is given by, } \]
\[I = \frac{2}{3}m r^2 + m l^2 \]
\[\text { Substituting the value of l in the above equation, we get: }\]
\[I = \frac{2}{3}m \left( 0 . 02 \right)^2 + m \left( 0 . 2 \right)^2 \]
\[ = \frac{2}{3}m\left( 0 . 0004 \right) + m\left( 0 . 04 \right)\]
\[ = m\left[ \frac{0 . 0008}{3} + 0 . 04 \right]\]
\[ = m\left( \frac{0 . 1208}{3} \right)\]
On substituting the value of I in equation (1) and differentiating it, we get:
\[\frac{d}{dt}\left[ mg \left( 0 . 2 \right) \left( 1 - \cos \theta \right) + \frac{1}{2}\frac{0 . 1208}{3}m \omega^2 \right] = \frac{d}{dt}\left( c \right)\]
\[ \Rightarrow mg\left( 0 . 2 \right)\sin\theta\frac{d\theta}{dt} + \frac{1}{2}\left( \frac{0 . 1208}{3} \right)m \times 2\omega\frac{d\omega}{dt} = 0 \]
\[ \Rightarrow 2\sin \theta = \frac{0 . 1208}{3}\alpha \left[ \text { because }, g = 10 m/ s^2 \right]\]
\[ \Rightarrow \frac{\alpha}{\theta} = \frac{6}{0 . 1208}\]
\[ \Rightarrow \omega^2 = 49 . 66\]
\[ \Rightarrow \omega = 7 . 04\]
\[\text { Thus, time period }\left( T \right) \text { will be: }\]
\[T = \frac{2\pi}{\omega} = 0 . 89 s\]
For a simple pendulum, time period (T) is given by,
% change in the value of time period = \[\frac{0 . 89 - 0 . 86}{0 . 89} \times 100 = 0 . 3\]
APPEARS IN
संबंधित प्रश्न
The average displacement over a period of S.H.M. is ______.
(A = amplitude of S.H.M.)
In a damped harmonic oscillator, periodic oscillations have _______ amplitude.
(A) gradually increasing
(B) suddenly increasing
(C) suddenly decreasing
(D) gradually decreasing
A pendulum clock gives correct time at the equator. Will it gain time or loose time as it is taken to the poles?
A block of known mass is suspended from a fixed support through a light spring. Can you find the time period of vertical oscillation only by measuring the extension of the spring when the block is in equilibrium?
A student says that he had applied a force \[F = - k\sqrt{x}\] on a particle and the particle moved in simple harmonic motion. He refuses to tell whether k is a constant or not. Assume that he was worked only with positive x and no other force acted on the particle.
The time period of a particle in simple harmonic motion is equal to the time between consecutive appearances of the particle at a particular point in its motion. This point is
The displacement of a particle in simple harmonic motion in one time period is
Figure represents two simple harmonic motions.
The parameter which has different values in the two motions is
A pendulum clock keeping correct time is taken to high altitudes,
A particle moves in the X-Y plane according to the equation \[\overrightarrow{r} = \left( \overrightarrow{i} + 2 \overrightarrow{j} \right)A\cos\omega t .\]
The motion of the particle is
(a) on a straight line
(b) on an ellipse
(c) periodic
(d) simple harmonic
A pendulum clock giving correct time at a place where g = 9.800 m/s2 is taken to another place where it loses 24 seconds during 24 hours. Find the value of g at this new place.
A simple pendulum of length 1 feet suspended from the ceiling of an elevator takes π/3 seconds to complete one oscillation. Find the acceleration of the elevator.
The length of a second’s pendulum on the surface of the Earth is 0.9 m. The length of the same pendulum on the surface of planet X such that the acceleration of the planet X is n times greater than the Earth is
A simple pendulum has a time period T1. When its point of suspension is moved vertically upwards according to as y = kt2, where y is the vertical distance covered and k = 1 ms−2, its time period becomes T2. Then, T `"T"_1^2/"T"_2^2` is (g = 10 ms−2)
Define the time period of simple harmonic motion.
A simple harmonic motion is given by, x = 2.4 sin ( 4πt). If distances are expressed in cm and time in seconds, the amplitude and frequency of S.H.M. are respectively,
A spring is stretched by 5 cm by a force of 10 N. The time period of the oscillations when a mass of 2 kg is suspended by it is ______.
A spring is stretched by 5 cm by a force of 10 N. The time period of the oscillations when a mass of 2 kg is suspended by it is ______
A body having specific charge 8 µC/g is resting on a frictionless plane at a distance 10 cm from the wall (as shown in the figure). It starts moving towards the wall when a uniform electric field of 100 V/m is applied horizontally toward the wall. If the collision of the body with the wall is perfectly elastic, then the time period of the motion will be ______ s.
