A body is executing simple harmonic motion of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) In simple harmonic motion,the velocity of the body is given by $v = \omega \sqrt{a^2 - x^2}$.The time $dt$ spent by the body in a small interval $

Vedclass · Accepted Answer

$x=\pm a$

Vedclass · Answer

(A) In simple harmonic motion,the velocity of the body is given by $v = \omega \sqrt{a^2 - x^2}$.The time $dt$ spent by the body in a small interval $dx$ is given by $dt = \frac{dx}{v} = \frac{dx}{\omega \sqrt{a^2 - x^2}}$.The probability $P(x)dx$ of finding the body in the interval $dx$ is proportional to the time spent in that interval,i.e.,$P(x) \propto \frac{1}{v} = \frac{1}{\omega \sqrt{a^2 - x^2}}$.As $x 	o \pm a$,the velocity $v 	o 0$,which means the time spent $dt 	o \infty$.Therefore,the probability of finding the body is highest at the extreme positions $x = \pm a$.Thus,the correct option is $A$.

$A$ body is executing simple harmonic motion of amplitude $a$ and period $T$ about the equilibrium position $x=0$. Large numbers of snapshots are taken at random of this body in motion. The probability of the body being found in a very small interval $x$ to $x+|dx|$ is highest at

Similar Questions

The displacement of a particle is represented by the equation $y = \sin^3 \omega t$. The motion is

What is the ratio between the distance travelled by the oscillator in one time period and the amplitude?

Draw a graph of displacement of $SHM$ as a function of time.

The circular motion of a particle with constant speed is

$A$ particle of mass $0.4 \text{ kg}$ executes simple harmonic motion of amplitude $0.4 \text{ m}$. When it passes through the mean position,its kinetic energy is $256 \times 10^{-3} \text{ J}$. If the initial phase of the oscillation is $\pi / 4$,then the equation of its motion is

Vedclass Test Series

Exam Paper Generator

Online Exam Module