Two particles $A$ and $B$ of equal masses are suspended from two massless springs of spring constants $k_1$ and $k_2$,respectively. If the maximum velocities during oscillations are equal,the ratio of amplitudes of $A$ and $B$ is

When a mass $m$ is hung from the lower end of a spring of negligible mass,an extension $x$ is produced in the spring. The time period of oscillation is

$A$ simple harmonic oscillator consists of a particle of mass $m$ and an ideal spring with spring constant $k$. The particle oscillates with a time period $T$. The spring is cut into two equal parts. If one part oscillates with the same particle, the new time period will be

$A$ $3 \ kg$ block is connected as shown in the figure. The spring constants of the two springs $K_1$ and $K_2$ are $50 \ Nm^{-1}$ and $150 \ Nm^{-1}$ respectively. The block is released from rest with the springs unstretched. The acceleration of the block in its lowest position is $(g=10 \ ms^{-2})$. (in $ms^{-2}$)

$A$ mass $m$ is vertically suspended from a spring of negligible mass; the system oscillates with a frequency $n$. What will be the frequency of the system if a mass $4m$ is suspended from the same spring?

The total mechanical energy of a spring-mass system in simple harmonic motion is $E = \frac{1}{2}m\omega^2 A^2$. Suppose the oscillating particle is replaced by another particle of double the mass while the amplitude $A$ remains the same. The new mechanical energy will: