A particle is executing SHM along a straight | vedclass

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

Question

(B) In $SHM$,the velocity $V$ of a particle at a distance $x$ from the mean position is given by $V = \omega \sqrt{a^2 - x^2}$,where $a$ is the amplit

Vedclass · Accepted Answer

$2\pi \sqrt {\frac{{{x_2}^2 - {x_1}^2}}{{{V_1}^2 - {V_2}^2}}}$

Vedclass · Answer

(B) In $SHM$,the velocity $V$ of a particle at a distance $x$ from the mean position is given by $V = \omega \sqrt{a^2 - x^2}$,where $a$ is the amplitude and $\omega$ is the angular frequency.Squaring both sides,we get $V^2 = \omega^2(a^2 - x^2)$.For distances $x_1$ and $x_2$,we have:$V_1^2 = \omega^2(a^2 - x_1^2) \dots (i)$$V_2^2 = \omega^2(a^2 - x_2^2) \dots (ii)$Subtracting equation $(ii)$ from $(i)$:$V_1^2 - V_2^2 = \omega^2(a^2 - x_1^2 - a^2 + x_2^2)$$V_1^2 - V_2^2 = \omega^2(x_2^2 - x_1^2)$$\omega^2 = \frac{V_1^2 - V_2^2}{x_2^2 - x_1^2}$Since the time period $T = \frac{2\pi}{\omega}$,we have $\omega = \frac{2\pi}{T}$.Substituting this into the equation for $\omega^2$:$\left(\frac{2\pi}{T}\right)^2 = \frac{V_1^2 - V_2^2}{x_2^2 - x_1^2}$$T^2 = 4\pi^2 \left(\frac{x_2^2 - x_1^2}{V_1^2 - V_2^2}\right)$$T = 2\pi \sqrt{\frac{x_2^2 - x_1^2}{V_1^2 - V_2^2}}$

$A$ particle is executing $SHM$ along a straight line. Its velocities at distances $x_1$ and $x_2$ from the mean position are $V_1$ and $V_2$ respectively. Its time period is

Similar Questions

The equations for the displacements of two particles in simple harmonic motion are $y_1=0.1 \sin \left(100 \pi t+\frac{\pi}{3}\right)$ and $y_2=0.1 \cos \pi t$ respectively. The phase difference between the velocities of the two particles at a time $t=0$ is

$A$ body is executing $S.H.M.$ When its displacement from the mean position is $4 \, cm$ and $5 \, cm$,the corresponding velocity of the body is $10 \, cm/sec$ and $8 \, cm/sec$. Then the time period of the body is

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

The displacement of a body executing $SHM$ is given by $x = A \sin (2\pi t + \pi /3).$ The first time from $t = 0$ when the velocity is maximum is .... $\sec$

Vedclass Test Series

Exam Paper Generator

Online Exam Module