English

Displacement versus time curve for a particle executing S.H.M. is shown in figure. Identify the points marked at which (i) velocity of the oscillator is zero, (ii) speed of the oscillator is maximum. - Physics

Advertisements
Advertisements

Question

Displacement versus time curve for a particle executing S.H.M. is shown in figure. Identify the points marked at which (i) velocity of the oscillator is zero, (ii) speed of the oscillator is maximum.

Short Note

Solution

In SHM y-t graph, zero displacement values correspond to the mean position; where the velocity of the oscillator is maximum.

Whereas the crest and troughs represent extreme positions, where displacement is maximum and velocity of the oscillator is minimum and is zero. Hence,

  1. A, C, E and G are either crests or trough having zero velocity.
  2. Speed is maximum at mean positions represented by B, D, F, and H paints.
shaalaa.com
Energy in Simple Harmonic Motion
  Is there an error in this question or solution?
Chapter 14: Oscillations - Exercises [Page 102]

APPEARS IN

NCERT Exemplar Physics [English] Class 11
Chapter 14 Oscillations
Exercises | Q 14.19 | Page 102

RELATED QUESTIONS

The maximum speed and acceleration of a particle executing simple harmonic motion are 10 cm/s and 50 cm/s2. Find the position(s) of the particle when the speed is 8 cm/s.


A block of mass 0.5 kg hanging from a vertical spring executes simple harmonic motion of amplitude 0.1 m and time period 0.314 s. Find the maximum force exerted by the spring on the block.


The spring shown in figure is unstretched when a man starts pulling on the cord. The mass of the block is M. If the man exerts a constant force F, find (a) the amplitude and the time period of the motion of the block, (b) the energy stored in the spring when the block passes through the equilibrium position and (c) the kinetic energy of the block at this position.


A 1 kg block is executing simple harmonic motion of amplitude 0.1 m on a smooth horizontal surface under the restoring force of a spring of spring constant 100 N/m. A block of mass 3 kg is gently placed on it at the instant it passes through the mean position. Assuming that the two blocks move together, find the frequency and the amplitude of the motion.


Show that for a particle executing simple harmonic motion.

  1. the average value of kinetic energy is equal to the average value of potential energy.
  2. average potential energy = average kinetic energy = `1/2` (total energy)

Hint: average kinetic energy = <kinetic energy> = `1/"T" int_0^"T" ("Kinetic energy") "dt"` and

average potential energy = <potential energy> = `1/"T" int_0^"T" ("Potential energy") "dt"`


When the displacement of a particle executing simple harmonic motion is half its amplitude, the ratio of its kinetic energy to potential energy is ______.


A body is executing simple harmonic motion with frequency ‘n’, the frequency of its potential energy is ______.


Find the displacement of a simple harmonic oscillator at which its P.E. is half of the maximum energy of the oscillator.


An object of mass 0.5 kg is executing a simple Harmonic motion. Its amplitude is 5 cm and the time period (T) is 0.2 s. What will be the potential energy of the object at an instant t = `T/4` s starting from the mean position? Assume that the initial phase of the oscillation is zero.


The total energy of a particle, executing simple harmonic motion is ______.

where x is the displacement from the mean position, hence total energy is independent of x.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×