Advertisements
Advertisements
Question
The time period of a particle in simple harmonic motion is equal to the smallest time between the particle acquiring a particular velocity \[\vec{v}\] . The value of v is
Options
vmax
0
between 0 and vmax
between 0 and −vmax
Solution
vmax
Because the time period of a simple harmonic motion is defined as the time taken to complete one oscillation.
APPEARS IN
RELATED QUESTIONS
A particle in S.H.M. has a period of 2 seconds and amplitude of 10 cm. Calculate the acceleration when it is at 4 cm from its positive extreme position.
Define phase of S.H.M.
Hence obtain the expression for acceleration, velocity and displacement of a particle performing linear S.H.M.
A particle executes simple harmonic motion. If you are told that its velocity at this instant is zero, can you say what is its displacement? If you are told that its velocity at this instant is maximum, can you say what is its displacement?
A particle executes simple harmonic motion Let P be a point near the mean position and Q be a point near an extreme. The speed of the particle at P is larger than the speed at Q. Still the particle crosses Pand Q equal number of times in a given time interval. Does it make you unhappy?
In measuring time period of a pendulum, it is advised to measure the time between consecutive passage through the mean position in the same direction. This is said to result in better accuracy than measuring time between consecutive passage through an extreme position. Explain.
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 average energy in one time period in simple harmonic motion is
For a particle executing simple harmonic motion, the acceleration is proportional to
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.
All the surfaces shown in figure are frictionless. The mass of the care is M, that of the block is m and the spring has spring constant k. Initially the car and the block are at rest and the spring is stretched through a length x0 when the system is released. (a) Find the amplitudes of the simple harmonic motion of the block and of the care as seen from the road. (b) Find the time period(s) of the two simple harmonic motions.
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.
A small block oscillates back and forth on a smooth concave surface of radius R ib Figure . Find the time period of small oscillation.
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.
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.
What is meant by simple harmonic oscillation? Give examples and explain why every simple harmonic motion is a periodic motion whereas the converse need not be true.
A body oscillates with SHM according to the equation x = 5 cos `(2π"t" + π/4)`. Its instantaneous displacement at t = 1 sec 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 ______
