$A$ force of $6.4 \,N$ stretches a vertical spring by $0.1 \,m$. If it were to oscillate with a period of $\frac{\pi}{4} \,s$, then the mass that is to be suspended from the spring is:

Two springs,of force constants $k_1$ and $k_2$ are connected to a mass $m$ as shown. The frequency of oscillation of the mass is $f$. If both $k_1$ and $k_2$ are made four times their original values,the frequency of oscillation becomes

Two masses $m_1$ and $m_2$ are suspended together by a massless spring of constant $K$. When the masses are in equilibrium,$m_1$ is removed without disturbing the system. The amplitude of oscillations is

On a frictionless horizontal plane,a bob of mass $m=0.1 \text{ kg}$ is attached to a spring with natural length $l_0=0.1 \text{ m}$. The spring constant is $k_1=0.009 \text{ N/m}$ when the length of the spring $l > l_0$ and is $k_2=0.016 \text{ N/m}$ when $l < l_0$. Initially,the bob is released from $l=0.15 \text{ m}$. Assume that Hooke's law remains valid throughout the motion. If the time period of the full oscillation is $T=(n \pi) \text{ s}$,then the integer closest to $n$ is:

$A$ particle connected to the end of a spring executes $S$.$H$.$M$. with period $T_1$. While the corresponding period for another spring is $T_2$. If the period of oscillation with two springs in series is $T$,then

$A$ block of mass $M_1$ is hung by a light spring of force constant $k$ from the top bar of a reverse $U$-frame of mass $M_2$ resting on the floor. The block is pulled down from its equilibrium position by a distance $x$ and then released. Find the minimum value of $x$ such that the reverse $U$-frame will leave the floor momentarily.