The displacement of a damped harmonic oscillator is given by $x\left(t\right)={\mathrm{e}}^{-0.1t} \cos \left(10\mathrm{\pi }t+\phi \right)$, where $t$ is the time in seconds. The time taken for the amplitude of the oscillator to drop to half of its initial value is close to

What will be the displacement equation of the simple harmonic motion obtained by combining the motions?
$x_1=2\sin \omega t, x_2=4\sin \left(\omega t+\frac{\pi }{6}\right) \mathrm{a}\mathrm{n}\mathrm{d} x_3=6\sin \left(\omega t+\frac{\pi }{3}\right)$

(Note: $\varphi$ is an acute angle)

A body of mass $0.01 \mathrm{kg}$ executes simple harmonic motion about $x=0$ under the influence of a force shown below. The period of the SHM is approximately

In the figure shown, the time period and the amplitude, respectively, when mass $m$ is released from rest and when the spring is relaxed are (The inclined plane is smooth)

A mass $m$ is suspended by means of two coiled springs which have the same length in unstretched condition, as shown in figure. Their force constants are $k_1$ and $k_2$, respectively. When set into vertical vibrations, the period will be

Which of the following is not correct about a  $20 \mathrm{g}$ particle subjected to two SHMs, $x_1=2\sin \left(10t\right)$, $x_2=4\sin \left(10t+\frac{\pi }{3}\right)$, where $x_1$ and $x_2$ are in metre and $t$ is in second.