The displacement of a damped harmonic oscillator is given by $x\left( t \right) = {e^{ - 0.1\,t}}\,\cos \left( {10\pi t + \varphi } \right)$ The time taken for its amplitude of vibration to drop to half of its initial value is close to .... $s$
JEE MAIN 2019, Medium
Download our app for free and get started
$A=A_{0} e^{-0.1 t}=\frac{A_{0}}{2}$
$\ln 2=0.1 \mathrm{t}$
$t=10\, \mathrm{ln} 2=6.93 \approx 7 \mathrm{sec}$
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
- 1A large horizontal surface moves up and down in $SHM$ with an amplitude of $1 \,cm$. If a mass of $10\, kg$ (which is placed on the surface) is to remain continually in contact with it, the maximum frequency of $S.H.M.$ will be ... $Hz$View Solution
- 2In the figure given below. a block of mass $M =490\,g$ placed on a frictionless table is connected with two springs having same spring constant $\left( K =2 N m ^{-1}\right)$. If the block is horizontally displaced through ' $X$ 'm then the number of complete oscillations it will make in $14 \pi$ seconds will be $.........$View Solution
- 3A circular disc of mass $10 \;kg$ is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be $1.5 \;s$. The radius of the disc is $15\; cm .$ Determine the torsional spring constant of the wire in $N\;m\;rad^{-1}$. (Torsional spring constant $\alpha$ is defined by the relation $J=-\alpha \theta,$ where $J$ is the restoring couple and $\theta$ the angle of twist).View Solution
- 4A $1.00 \times {10^{ - 20}}kg$ particle is vibrating with simple harmonic motion with a period of $1.00 \times {10^{ - 5}}sec$ and a maximum speed of $1.00 \times {10^3}m/s$. The maximum displacement of the particle isView Solution
- 5The 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
- 6Assume that the earth is a solid sphere of uniform density and a tunnel is dug along its diameter throughout the earth. It is found that when a particle is released in this tunnel, it executes a simple harmonic motion. The mass of the particle is $100 g$. The time period of the motion of the particle will be (approximately) (take $g =10\,ms ^{-2}$, radius of earth $=6400\,km$ )View Solution
- 7${T}_{0}$ is the time period of a simple pendulum at a place. If the length of the pendulum is reduced to $\frac{1}{16}$ times of its initial value, the modified timeView Solution
- 8A particle of mass $4 \,kg$ moves simple harmonically such that its $P E(U)$ varies with position $x$, as shown. The period of oscillations is ............View Solution
- 9The displacement of a particle varies according to the relation $x = 3 \ sin\ 100t + 8\ cos^2\ 50t$ . Which of the following is incorrect about this motionView Solution
- 10The resultant of two rectangular simple harmonic motions of the same frequency and equal amplitudes but differing in phase by $\frac{\pi }{2}$ isView Solution