A particle is making simple harmonic motion a | vedclass

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

Question

(B) The velocity $v$ of a particle in simple harmonic motion at a displacement $x$ is given by $v^{2} = \omega^{2}(A^{2} - x^{2})$,where $A$ is the am

Vedclass · Accepted Answer

$T=2 \pi \sqrt{\frac{x_{2}^{2}-x_{1}^{2}}{v_{1}^{2}-v_{2}^{2}}}$

Vedclass · Answer

(B) The velocity $v$ of a particle in simple harmonic motion at a displacement $x$ is given by $v^{2} = \omega^{2}(A^{2} - x^{2})$,where $A$ is the amplitude and $\omega$ is the angular frequency.For the given positions:$v_{1}^{2} = \omega^{2}(A^{2} - x_{1}^{2}) \implies \frac{v_{1}^{2}}{\omega^{2}} = A^{2} - x_{1}^{2} \implies A^{2} = x_{1}^{2} + \frac{v_{1}^{2}}{\omega^{2}}$$v_{2}^{2} = \omega^{2}(A^{2} - x_{2}^{2}) \implies \frac{v_{2}^{2}}{\omega^{2}} = A^{2} - x_{2}^{2} \implies A^{2} = x_{2}^{2} + \frac{v_{2}^{2}}{\omega^{2}}$Equating the two expressions for $A^{2}$:$x_{1}^{2} + \frac{v_{1}^{2}}{\omega^{2}} = x_{2}^{2} + \frac{v_{2}^{2}}{\omega^{2}}$$\frac{v_{1}^{2} - v_{2}^{2}}{\omega^{2}} = x_{2}^{2} - x_{1}^{2}$$\omega^{2} = \frac{v_{1}^{2} - v_{2}^{2}}{x_{2}^{2} - x_{1}^{2}}$Since $T = \frac{2\pi}{\omega}$,we have $\omega^{2} = \frac{4\pi^{2}}{T^{2}}$.$\frac{4\pi^{2}}{T^{2}} = \frac{v_{1}^{2} - v_{2}^{2}}{x_{2}^{2} - x_{1}^{2}}$$T^{2} = 4\pi^{2} \frac{x_{2}^{2} - x_{1}^{2}}{v_{1}^{2} - v_{2}^{2}}$$T = 2\pi \sqrt{\frac{x_{2}^{2} - x_{1}^{2}}{v_{1}^{2} - v_{2}^{2}}}$

$A$ particle is making simple harmonic motion along the $x$-axis. If at a distance $x_{1}$ and $x_{2}$ from the mean position the velocities of the particle are $v_{1}$ and $v_{2}$ respectively,the time period of its oscillation is given as -

Similar Questions

Two simple harmonic motions are represented by equations $y_1 = 4 \sin(10t + \phi)$ and $y_2 = 5 \cos(10t)$. What is the phase difference between their velocities?

The equation of a particle executing simple harmonic motion is given by $x = \sin \pi (t + 1/3) \, m$. At $t = 1 \, s$,the speed of the particle will be .......... $cm \, s^{-1}$ (Given: $\pi = 3.14$)

$A$ particle is executing simple harmonic motion in one-dimension. If the amplitude of oscillations is $0.2 \,cm$ and if its velocity at the mean position is $5 \,m/s$, then the angular frequency of the oscillation is

The maximum velocity of a particle performing $S.H.M.$ is $V$. If the periodic time is made $\left(\frac{1}{3}\right)^{rd}$ of its original value and the amplitude is doubled,then the new maximum velocity of the particle will be:

$A$ $1.00 \times 10^{-20} \,kg$ particle is vibrating under simple harmonic motion with a period of $1.00 \times 10^{-5} \,s$ and with a maximum speed of $1.00 \times 10^3 \,m/s$. The maximum displacement of the particle from the mean position is .......... $mm$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module