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
A body of mass 1 kg is made to oscillate on a spring of force constant 16 N/m. Calculate:
a) Angular frequency
b) frequency of vibration.
A particle executing simple harmonic motion comes to rest at the extreme positions. Is the resultant force on the particle zero at these positions according to Newton's first law?
The energy of system in simple harmonic motion is given by \[E = \frac{1}{2}m \omega^2 A^2 .\] Which of the following two statements is more appropriate?
(A) The energy is increased because the amplitude is increased.
(B) The amplitude is increased because the energy is increased.
A hollow sphere filled with water is used as the bob of a pendulum. Assume that the equation for simple pendulum is valid with the distance between the point of suspension and centre of mass of the bob acting as the effective length of the pendulum. If water slowly leaks out of the bob, how will the time period vary?
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 displacement of a particle in simple harmonic motion in one time period is
Suppose a tunnel is dug along a diameter of the earth. A particle is dropped from a point, a distance h directly above the tunnel. The motion of the particle as seen from the earth is
(a) simple harmonic
(b) parabolic
(c) on a straight line
(d) periodic
A particle moves on the X-axis according to the equation x = x0 sin2 ωt. The motion is simple harmonic
In a simple harmonic motion
The angle made by the string of a simple pendulum with the vertical depends on time as \[\theta = \frac{\pi}{90} \sin \left[ \left( \pi s^{- 1} \right)t \right]\] .Find the length of the pendulum if g = π2 m2.
The pendulum of a certain clock has time period 2.04 s. How fast or slow does the clock run during 24 hours?
A simple pendulum is constructed by hanging a heavy ball by a 5.0 m long string. It undergoes small oscillations. (a) How many oscillations does it make per second? (b) What will be the frequency if the system is taken on the moon where acceleration due to gravitation of the moon is 1.67 m/s2?
A simple pendulum of length 40 cm is taken inside a deep mine. Assume for the time being that the mine is 1600 km deep. Calculate the time period of the pendulum there. Radius of the earth = 6400 km.
Assume that a tunnel is dug across the earth (radius = R) passing through its centre. Find the time a particle takes to cover the length of the tunnel if (a) it is projected into the tunnel with a speed of \[\sqrt{gR}\] (b) it is released from a height R above the tunnel (c) it is thrown vertically upward along the length of tunnel with a speed of \[\sqrt{gR}\]
A simple pendulum of length l is suspended through the ceiling of an elevator. Find the time period of small oscillations if the elevator (a) is going up with and acceleration a0(b) is going down with an acceleration a0 and (c) is moving with a uniform velocity.
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
The displacement of a particle varies with time according to the relation y = a sin ωt + b cos ωt.
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.
Assume there are two identical simple pendulum clocks. Clock - 1 is placed on the earth and Clock - 2 is placed on a space station located at a height h above the earth's surface. Clock - 1 and Clock - 2 operate at time periods 4 s and 6 s respectively. Then the value of h is ______.
(consider the radius of earth RE = 6400 km and g on earth 10 m/s2)
