The amplitude and the time period in a $S.H.M.$ is $0.5 \,cm$ and $0.4 \,sec$ respectively. If the initial phase is $\pi /2$ radian, then the equation of $S.H.M.$ will be
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Two bodies performing $SHM$ have same amplitude and frequency. Their phases at a certain instant are as shown in the figure. The phase difference between them is
A mass $M$ is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes simple harmonic oscillations with a time period $T$. If the mass is increased by m then the time period becomes $\left( {\frac{5}{4}T} \right)$. The ratio of $\frac{m}{{M}}$ is
The displacements of two particles executing $S.H.M.$ on the same line are given. as $y_1=a \sin \left(\frac{\pi}{2} t+\phi\right)$ and $y_2=b \sin \left(\frac{2 \pi}{3} t+\phi\right)$. The phase difference between them at $t=1 \,s$ is .........
Two pendulum have time periods $T$ and $5T/4$. They start $SHM$ at the same time from the mean position. After how many oscillations of the smaller pendulum they will be again in the same phase
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
For a particle executing simple harmonic motion, the kinetic energy $K$ is given by $K = {K_o}{\cos ^2}\omega t$. The maximum value of potential energy is
A simple pendulum with a bob of mass $‘m’$ oscillates from $A$ to $ C$ and back to $A$ such that $PB$ is $H.$ If the acceleration due to gravity is $‘g’$, then the velocity of the bob as it passes through $B$ is
Two pendulums begin to swing simultaneously. If the ratio of the frequency of oscillations of the two is $7 : 8$, then the ratio of lengths of the two pendulums will be