Three masses $700 \, g, 500 \, g$,and $400 \, g$ are suspended at the end of a spring as shown and are in equilibrium. When the $700 \, g$ mass is removed,the system oscillates with a period of $3 \, s$. When the $500 \, g$ mass is also removed,it will oscillate with a period of ...... $s$.

$A$ mass $M$ attached to a horizontal spring executes simple harmonic motion with amplitude $A_1$. When mass $M$ passes the mean position,a smaller mass $m$ is attached to it,and both of them together execute simple harmonic motion with amplitude $A_2$. Then the value of $\frac{A_1}{A_2}$ is

$A$ silver atom in a solid oscillates in simple harmonic motion in a specific direction with a frequency of $10^{12} \ s^{-1}$. What is the force constant of the bonds connecting one atom to the others? (Molar mass of silver $= 108 \ g/mol$ and Avogadro number $= 6.02 \times 10^{23} \ mol^{-1}$)

As per the given figures,two springs of spring constants $K$ and $2K$ are connected to a mass $m$. If the period of oscillation in figure $(a)$ is $3 \text{ s}$,then the period of oscillation in figure $(b)$ will be $\sqrt{x} \text{ s}$. The value of $x$ is $.........$

$A$ body of mass $1 \,kg$ is suspended from a spring of negligible mass. Another body of mass $500 \,g$ moving vertically upwards hits the suspended body with a velocity of $3 \,ms^{-1}$ and gets embedded in it. If the frequency of oscillation of the system of the two bodies after collision is $\frac{10}{\pi} \,Hz$,the amplitude of the motion and the spring constant are respectively,

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: