A $S.H.M.$ has amplitude $‘a’$ and time period $T$. The maximum velocity will be
Easy
Download our app for free and get started
(d) ${v_{\max }} = a\omega = \frac{{a\,.\,2\pi }}{T} = \frac{{2\pi a}}{T}$
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1If displacement $x$ and velocity $v$ are related as $4v^2 = 25 -x^2$ in a $SHM$. Then time Period of given $SHM$ is (Consider $SI\, units$)View Solution
- 2The amplitude of vibration of a particle is given by ${a_m} = ({a_0})/(a{\omega ^2} - b\omega + c);$ where ${a_0},a,b$ and $c$ are positive. The condition for a single resonant frequency isView Solution
- 3The displacement of an object attached to a spring and executing simple harmonic motion is given by $ x= 2 \times 10^{-9}$ $ cos$ $\;\pi t\left( m \right)$ .The time at which the maximum speed first occurs isView Solution
- 4The resultant of two rectangular simple harmonic motions of the same frequency and equal amplitudes but differing in phase by $\frac{\pi }{2}$ isView Solution
- 5A simple pendulum has time period $T$. The bob is given negative charge and surface below it is given positive charge. The new time period will beView Solution
- 6The kinetic energy and potential energy of a particle executing simple harmonic motion will be equal, when displacement (amplitude = $a$) isView Solution
- 7A function is represented by equationView Solution
$y = A\,\cos \,\omega t\,\cos \,2\omega t + A\,\sin \,\omega t\,\sin \,2\omega t$.
Than the nature of the function is
- 8A large horizontal surface moves up and down in $S.H.M.$ with an amplitude of $1\, cm$. If a mass of $10\, kg$ (which is placed on the surface is to remain continuously in contact with it, the maximum frequency of $S.H.M.$ will be .... $Hz$View Solution
- 9A mass $m$ is attached to two springs as shown in figure. The spring constants of two springs are $K _1$ and $K _2$. For the frictionless surface, the time period of oscillation of mass $m$ isView Solution
- 10A particle executes $S.H.M.$ of amplitude A along $x$-axis. At $t =0$, the position of the particle is $x=\frac{A}{2}$ and it moves along positive $x$-axis the displacement of particle in time $t$ is $x=A \sin (\omega t+\delta)$, then the value $\delta$ will beView Solution