The kinetic energy,K of a body performing sim | vedclass

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

Question

(A) The kinetic energy $K$ of a body performing simple harmonic motion is given by the formula:$K = \frac{1}{2} m v^2$where $v$ is the velocity of the

Vedclass · Accepted Answer

Vedclass · Answer

(A) The kinetic energy $K$ of a body performing simple harmonic motion is given by the formula:$K = \frac{1}{2} m v^2$where $v$ is the velocity of the body.If the displacement is $y = a \sin \omega t$,then the velocity is:$v = \frac{dy}{dt} = a \omega \cos \omega t$Substituting this into the kinetic energy expression:$K = \frac{1}{2} m (a \omega \cos \omega t)^2 = \frac{1}{2} m a^2 \omega^2 \cos^2 \omega t$Using the trigonometric identity $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$,we get:$K = \frac{1}{2} m a^2 \omega^2 \left( \frac{1 + \cos 2\omega t}{2} ight) = \frac{1}{4} m a^2 \omega^2 (1 + \cos 2\omega t)$This expression shows that the kinetic energy $K$ is always non-negative and varies with a frequency double that of the simple harmonic motion (i.e.,frequency $2\omega$). The graph in option $A$ correctly represents this periodic variation,where $K$ oscillates between $0$ and a maximum value with a period of $T/2$.

The kinetic energy,$K$ of a body performing simple harmonic motion varies with time $t$,is indicated in graph

Similar Questions

The total energy of a particle executing $S.H.M.$ is $80 \, J$. What is the potential energy when the particle is at a distance of $\frac{3}{4}$ of amplitude from the mean position?

$A$ particle starts simple harmonic motion from the mean position. Its amplitude is $a$ and total energy is $E$. At one instant,its kinetic energy is $3E/4$. Its displacement at that instant is:

The position of a particle executing simple harmonic motion is given by $x(t) = 2 \cos \left(\frac{\pi}{15} t - \frac{\pi}{2}\right)$, where $x$ is in centimetre and $t$ is in seconds. The time period of the kinetic energy of the particle in seconds is

The ratio between kinetic and potential energies of a body executing simple harmonic motion, when it is at a distance of $\frac{1}{N}$ of its amplitude from the mean position is

$A$ mass of $1 \text{ kg}$ is executing $SHM$. Its displacement is given by $x = 6.0 \cos(100t + \pi/4) \text{ cm}$. What is the maximum kinetic energy (in $text{ J}$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module