Advertisements
Advertisements
प्रश्न
In a simple harmonic motion
पर्याय
the potential energy is always equal to the kinetic energy
the potential energy is never equal to the kinetic energy
the average potential energy in any time interval is equal to the average kinetic energy in that time interval
the average potential energy in one time period is equal to the average kinetic energy in this period.
Advertisements
उत्तर
the average potential energy in one time period is equal to the average kinetic energy in this period.
The kinetic energy of the motion is given as,
\[\frac{1}{2}k A^2 \cos^2 \omega t\]
The potential energy is calculated as,
APPEARS IN
संबंधित प्रश्न
Define phase of S.H.M.
Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion?
(a) a = 0.7x
(b) a = –200x2
(c) a = –10x
(d) a = 100x3
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.
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.
Hence obtain the expression for acceleration, velocity and displacement of a particle performing linear S.H.M.
A pendulum clock gives correct time at the equator. Will it gain time or loose time as it is taken to the poles?
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
The average energy in one time period in simple harmonic motion is
A pendulum clock that keeps correct time on the earth is taken to the moon. It will run
A particle moves in a circular path with a continuously increasing speed. Its motion is
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 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 along a chord of the earth, at a perpendicular distance R/2 from the earth's centre where R is the radius of the earth. The wall of the tunnel is frictionless. (a) Find the gravitational force exerted by the earth on a particle of mass mplaced in the tunnel at a distance x from the centre of the tunnel. (b) Find the component of this force along the tunnel and perpendicular to the tunnel. (c) Find the normal force exerted by the wall on the particle. (d) Find the resultant force on the particle. (e) Show that the motion of the particle in the tunnel is simple harmonic and find the time period.
A simple pendulum fixed in a car has a time period of 4 seconds when the car is moving uniformly on a horizontal road. When the accelerator is pressed, the time period changes to 3.99 seconds. Making an approximate analysis, find the acceleration of the car.
Three simple harmonic motions of equal amplitude A and equal time periods in the same direction combine. The phase of the second motion is 60° ahead of the first and the phase of the third motion is 60° ahead of the second. Find the amplitude of the resultant motion.
A particle is subjected to two simple harmonic motions, one along the X-axis and the other on a line making an angle of 45° with the X-axis. The two motions are given by x = x0 sin ωt and s = s0 sin ωt. Find the amplitude of the resultant motion.
State the laws of the simple pendulum?
Motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower point is ______.
- simple harmonic motion.
- non-periodic motion.
- periodic motion.
- periodic but not S.H.M.
A weightless rigid rod with a small iron bob at the end is hinged at point A to the wall so that it can rotate in all directions. The rod is kept in the horizontal position by a vertical inextensible string of length 20 cm, fixed at its midpoint. The bob is displaced slightly, perpendicular to the plane of the rod and string. The period of small oscillations of the system in the form `(pix)/10` is ______ sec. and the value of x is ______.
(g = 10 m/s2)
