The general displacement of a simple harmonic | vedclass

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

Question

(D) The displacement is given by $x = A \sin \omega t$.The potential energy $U$ is given by $U = \frac{1}{2} k x^2 = \frac{1}{2} k A^2 \sin^2 \omega t

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) The displacement is given by $x = A \sin \omega t$.The potential energy $U$ is given by $U = \frac{1}{2} k x^2 = \frac{1}{2} k A^2 \sin^2 \omega t$.The slope of the potential energy-time curve is $\frac{dU}{dt}$.$\frac{dU}{dt} = \frac{d}{dt} (\frac{1}{2} k A^2 \sin^2 \omega t) = \frac{1}{2} k A^2 (2 \sin \omega t \cos \omega t) \cdot \omega = \frac{1}{2} k A^2 \omega \sin(2 \omega t)$.For the slope to be maximum,$\sin(2 \omega t)$ must be maximum,i.e.,$\sin(2 \omega t) = 1$.This occurs when $2 \omega t = \frac{\pi}{2}$.Substituting $\omega = \frac{2 \pi}{T}$,we get $2 (\frac{2 \pi}{T}) t = \frac{\pi}{2}$.$\frac{4 \pi t}{T} = \frac{\pi}{2} \implies t = \frac{T}{8}$.Comparing this with $t = \frac{T}{\beta}$,we get $\beta = 8$.

The general displacement of a simple harmonic oscillator is $x = A \sin \omega t$. Let $T$ be its time period. The slope of its potential energy $(U)$ - time $(t)$ curve will be maximum when $t = \frac{T}{\beta}$. The value of $\beta$ is $.........$

Similar Questions

The $P.E.$ of a particle executing $SHM$ at a distance $x$ from its equilibrium position is

The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (where $E$ is the total energy)

$A$ body performs linear simple harmonic motion of amplitude $A$. At what displacement from the mean position is the potential energy of the body one-fourth of its total energy?

The potential energy of a simple harmonic oscillator of mass $2 \,kg$ at its mean position is $5 \,J$. If its total energy is $9 \,J$ and amplitude is $1 \,cm$, then its time period is

$A$ particle starts oscillating simple harmonically from its equilibrium position with time period $T$. What is the ratio of potential energy to kinetic energy of the particle at time $t = \frac{T}{12}$? (Given: $\sin(\frac{\pi}{6}) = \frac{1}{2}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module