In damped oscillations, damping force is directly proportional to speed of oscillator. If amplitude becomes half of its maximum value in $1 \,s$, then after $2 \,s$ amplitude will be $\left(A_0-\right.$ initial amplitude)
Medium
Download our app for free and get started
(a)
$A=A_0 e^{-b t}$
Amplitude becomes half hence
$\frac{A_0}{2} A_0 e^{-b t}[t=1]$
$\therefore e^{-b}=\frac{1}{2}$
$\therefore \text { In two seconds }$
$A=A_0\left(\frac{1}{2}\right)^2$
$A=\frac{A_0}{4}$
Download our app
and 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.*
In the reported figure, two bodies $A$ and $B$ of masses $200\, {g}$ and $800\, {g}$ are attached with the system of springs. Springs are kept in a stretched position with some extension when the system is released. The horizontal surface is assumed to be frictionless. The angular frequency will be $.....\,{rad} / {s}$ when ${k}=20 \,{N} / {m} .$
A particle performs $SHM$ about $x = 0$ such that at $t = 0$ it is at $x = 0$ and moving towards positive extreme. The time taken by it to go from $x = 0$ to $x = \frac{A}{2}$ is ..... times the time taken to go from $x = \frac{A}{2}$ to $A$. The most suitable option for the blank space is
A block of mass $2\,kg$ is attached with two identical springs of spring constant $20\,N / m$ each. The block is placed on a frictionless surface and the ends of the springs are attached to rigid supports (see figure). When the mass is displaced from its equilibrium position, it executes a simple harmonic motion. The time period of oscillation is $\frac{\pi}{\sqrt{x}}$ in SI unit. The value of $x$ is $..........$
If a Second's pendulum is moved to a planet where acceleration due to gravity is $4$ times, the length of the second's pendulum on the planet should be made .......... times