Advertisements
Advertisements
Question
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.
Advertisements
Solution
It is given that:
Angle made by the simple pendulum with the vertical, \[ \theta = \left( \frac{\pi}{90} \right)\sin \left[ \pi\left( s^{- 1} \right)t \right]\] On comparing the above equation with the equation of S . H . M . , we get:
\[\omega = \pi s^{- 1} \]
\[ \Rightarrow \frac{2\pi}{T} = \pi\]
\[ \therefore T = 2 s\]
\[\text { Time period is given by the relation, }\]
\[T = 2\pi\sqrt{\left( \frac{l}{g} \right)}\]
\[ \Rightarrow 2 = 2\pi\sqrt{\left( \frac{l}{\pi^2} \right)}\]
\[ \Rightarrow 1 = \pi\frac{1}{\pi}\sqrt{\left( l \right)}\]
\[ \Rightarrow l = 1 m\]
\[\text { Hence, length of the pendulum is 1 m .}\]
APPEARS IN
RELATED QUESTIONS
In a damped harmonic oscillator, periodic oscillations have _______ amplitude.
(A) gradually increasing
(B) suddenly increasing
(C) suddenly decreasing
(D) gradually decreasing
Assuming the expression for displacement of a particle starting from extreme position, explain graphically the variation of velocity and acceleration w.r.t. time.
A particle executes S.H.M. with a period of 10 seconds. Find the time in which its potential energy will be half of its total energy.
Can a pendulum clock be used in an earth-satellite?
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?
The displacement of a particle in simple harmonic motion in one time period is
The displacement of a particle is given by \[\overrightarrow{r} = A\left( \overrightarrow{i} \cos\omega t + \overrightarrow{j} \sin\omega t \right) .\] The motion of the particle is
A pendulum clock that keeps correct time on the earth is taken to the moon. It will run
Which of the following will change the time period as they are taken to moon?
(a) A simple pendulum
(b) A physical pendulum
(c) A torsional pendulum
(d) A spring-mass system
A particle executes simple harmonic motion with an amplitude of 10 cm and time period 6 s. At t = 0 it is at position x = 5 cm going towards positive x-direction. Write the equation for the displacement x at time t. Find the magnitude of the acceleration of the particle at t = 4 s.
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 small block oscillates back and forth on a smooth concave surface of radius R ib Figure . Find the time period of small oscillation.
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?
A particle is subjected to two simple harmonic motions of same time period in the same direction. The amplitude of the first motion is 3.0 cm and that of the second is 4.0 cm. Find the resultant amplitude if the phase difference between the motions is (a) 0°, (b) 60°, (c) 90°.
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
Write short notes on two springs connected in parallel.
Consider the Earth as a homogeneous sphere of radius R and a straight hole is bored in it through its centre. Show that a particle dropped into the hole will execute a simple harmonic motion such that its time period is
T = `2π sqrt("R"/"g")`
The displacement of a particle is represented by the equation y = sin3ωt. The motion is ______.
Displacement vs. time curve for a particle executing S.H.M. is shown in figure. Choose the correct statements.

- Phase of the oscillator is same at t = 0 s and t = 2s.
- Phase of the oscillator is same at t = 2 s and t = 6s.
- Phase of the oscillator is same at t = 1 s and t = 7s.
- Phase of the oscillator is same at t = 1 s and t = 5s.
The velocities of a particle in SHM at positions x1 and x2 are v1 and v2 respectively, its time period will be ______.
