Advertisements
Advertisements
Question
In a simple harmonic motion
Options
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
Solution
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
RELATED QUESTIONS
In a damped harmonic oscillator, periodic oscillations have _______ amplitude.
(A) gradually increasing
(B) suddenly increasing
(C) suddenly decreasing
(D) gradually decreasing
Hence obtain the expression for acceleration, velocity and displacement of a particle performing linear S.H.M.
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?
Can the potential energy in a simple harmonic motion be negative? Will it be so if we choose zero potential energy at some point other than the mean position?
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 motion of a particle is given by x = A sin ωt + B cos ωt. The motion of the particle is
A wall clock uses a vertical spring-mass system to measure the time. Each time the mass reaches an extreme position, the clock advances by a second. The clock gives correct time at the equator. If the clock is taken to the poles it will
The motion of a torsional pendulum is
(a) periodic
(b) oscillatory
(c) simple harmonic
(d) angular simple harmonic
The pendulum of a certain clock has time period 2.04 s. How fast or slow does the clock run during 24 hours?
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 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°.
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)
If the inertial mass and gravitational mass of the simple pendulum of length l are not equal, then the time period of the simple pendulum is
State the laws of the simple pendulum?
Describe Simple Harmonic Motion as a projection of uniform circular motion.
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.
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)
