A particle executes $SHM$ with amplitude of $20 \,cm$ and time period is $12\, sec$.  What is the minimum time required for it to move between two points $10\, cm$ on  either side of the mean position ..... $\sec$ ?

A mass $m$ is suspended from the two coupled springs connected in series. The force constant for springs are ${K_1}$ and ${K_2}$. The time period of the suspended mass will be

A mass m oscillates with simple harmonic motion with frequency $f = \frac{\omega }{{2\pi }}$ and amplitude A on a spring with constant $K$ , therefore

Initially system is in equilibrium. Time period of $SHM$ of block in vertical direction is

A mass $m =100\, gms$ is attached at the end of a light spring which oscillates on a frictionless horizontal table with an amplitude equal to $0.16$ metre and time period equal to $2 \,sec$. Initially the mass is released from rest at $t = 0$ and displacement $x = - 0.16$ metre. The expression for the displacement of the mass at any time $t$ is

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 $4 m$ is suspended from the same spring