The position of a particle executing simple h | vedclass

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

Question

(C) The given equation for position in $SHM$ is $x(t) = 2 \cos \left(\frac{\pi}{15} t - \frac{\pi}{2}\right)$.Comparing this with the standard equatio

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(C) The given equation for position in $SHM$ is $x(t) = 2 \cos \left(\frac{\pi}{15} t - \frac{\pi}{2}ight)$.Comparing this with the standard equation $x(t) = A \cos(\omega t + \phi)$,we get the angular frequency $\omega = \frac{\pi}{15} 	ext{ rad/s}$.The time period of the $SHM$ is given by $T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi/15} = 30 	ext{ s}$.The kinetic energy of a particle in $SHM$ is given by $KE = \frac{1}{2} m v^2 = \frac{1}{2} m \omega^2 A^2 \sin^2(\omega t + \phi)$.Using the identity $\sin^2 	heta = \frac{1 - \cos(2	heta)}{2}$,the kinetic energy varies with a frequency of $2\omega$.Therefore,the time period of the kinetic energy is $T_{KE} = \frac{T}{2} = \frac{30}{2} = 15 	ext{ s}$.

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

Similar Questions

Consider the following statements. The total energy of a particle executing simple harmonic motion depends on its:
$(1)$ Amplitude $(2)$ Period $(3)$ Displacement
Of these statements:

In $SHM$,the restoring force is $F = -kx$,where $k$ is the force constant,$x$ is the displacement,and $A$ is the amplitude of motion. Then,the total energy depends upon:

$A$ simple pendulum of mass $m$ executes $S.H.M.$ with total energy $E$. If at an instant it is at one of the extreme positions,then its linear momentum after a phase shift of $\frac{\pi}{3} \, rad$ will be

Write the coordinates of the points of intersection of the graph of kinetic energy $(K)$ and potential energy $(U)$ for a simple harmonic oscillator.

$A$ particle starts from the mean position and performs $S.H.M.$ with a period of $6 \ s$. At what time is its kinetic energy $50 \%$ of its total energy (in $s$)? $\left(\cos 45^{\circ} = \frac{1}{\sqrt{2}}\right)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module