The motion of a mass on a spring, with spring constant ${K}$ is as shown in figure. The equation of motion is given by $x(t)= A sin \omega t+ Bcos\omega t$ with $\omega=\sqrt{\frac{K}{m}}$ Suppose that at time $t=0$, the position of mass is $x(0)$ and velocity $v(0)$, then its displacement can also be represented as $x(t)=C \cos (\omega t-\phi)$, where $C$ and $\phi$ are
${C}=\sqrt{\frac{2 {v}(0)^{2}}{\omega^{2}}+{x}(0)^{2}}, \phi=\tan ^{-1}\left(\frac{{x}(0) \omega}{2 {v}(0)}\right)$
${C}=\sqrt{\frac{{v}(0)^{2}}{\omega^{2}}+{x}(0)^{2}}, \phi=\tan ^{-1}\left(\frac{{x}(0) \omega}{{v}(0)}\right)$
$C=\sqrt{\frac{2 v(0)^{2}}{\omega^{2}}+x(0)^{2}}, \phi=\tan ^{-1}\left(\frac{v(0)}{x(0) \omega}\right)$
${C}=\sqrt{\frac{{v}(0)^{2}}{\omega^{2}}+{x}(0)^{2}}, \phi=\tan ^{-1}\left(\frac{{v}(0)}{{x}(0) \omega}\right)$
As per given figures, two springs of spring constants $K$ and $2\,K$ are connected to mass $m$. If the period of oscillation in figure $(a)$ is $3 s$, then the period of oscillation in figure $(b)$ will be $\sqrt{ x }$ s. The value of $x$ is$.........$
How the period of oscillation depend on the mass of block attached to the end of spring ?
Two masses $m_1$ and $m_2$ are suspended together by a massless spring of constant $K$. When the masses are in equilibrium, $m_1$ is removed without disturbing the system. The amplitude of oscillations is
If a spring extends by $x$ on loading, then energy stored by the spring is (if $T$ is the tension in the spring and $K$ is the spring constant)
A spring has spring constant $k$ and $l$. If it cut into piece spring in the proportional to $\alpha : \beta : \gamma $ then obtain the spring constant of every piece in term of spring constant of original spring (Here, $\alpha $, $\beta $ and $\gamma $ are integers)