Springs of spring constants $K, 2K, 4K, 8K,$ ..... are connected in series. A mass $40\, gm$ is attached to the lower end of last spring and the system is allowed to vibrate. What is the time period of oscillation ..... $\sec$. (Given $K = 2\, N/cm$)
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A vertical mass spring system executes simple harmonic oscillations with a period of $2\,s$. A quantity of this system which exhibits simple harmonic variation with a period of $1\, sec$ is
Equations ${y_1} = A\sin \omega t$ and ${y_2} = \frac{A}{2}\sin \omega t + \frac{A}{2}\cos \omega t$ represent $S.H.M.$ The ratio of the amplitudes of the two motions is
A particle executes simple harmonic motion with an amplitude of $4 \mathrm{~cm}$. At the mean position, velocity of the particle is $10 \mathrm{~cm} / \mathrm{s}$. The distance of the particle from the mean position when its speed becomes $5 \mathrm{~cm} / \mathrm{s}$ is $\sqrt{\alpha} \mathrm{cm}$, where $\alpha=$____________.
The amplitude of a damped oscillator becomes one third in $2\, sec$. If its amplitude after $6\, sec$ is $1/n$ times the original amplitude then the value of $n$ is
When an oscillator completes $100$ oscillations its amplitude reduced to $\frac{1}{3}$ of initial value. What will be its amplitude, when it completes $200$ oscillations?
For a particle executing $S.H.M.,\, x =$ displacement from equilibrium position, $v =$ velocity at any instant and $a =$ acceleration at any instant, then
A particle executes linear simple harmonic motion with an amplitude of $2\, cm$. When the particle is at $1\, cm$ from the mean position the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is
A particle of mass $m$ oscillates with simple harmonic motion between points ${x_1}$ and ${x_2}$, the equilibrium position being $O$. Its potential energy is plotted. It will be as given below in the graph