In the arrangement given in the figure,if the block of mass $m$ is displaced,the frequency is given by:

$A$ mass $m$ is suspended from a spring of length $l$ and force constant $K$. The frequency of vibration of the mass is $f_1$. The spring is cut into two equal parts and the same mass is suspended from one of the parts. The new frequency of vibration of the mass is $f_2$. Which of the following relations between the frequencies is correct?

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 vibration is

$A$ particle of mass $1 \, \text{kg}$ is hanging from a spring of force constant $100 \, \text{N/m}$. The mass is pulled slightly downward and released so that it executes free simple harmonic motion with time period $T$. The time when the kinetic energy and potential energy of the system will become equal is $\frac{T}{x}$. The value of $x$ is ..... .

$A$ spring with a spring constant $1200 \; N m^{-1}$ is mounted on a horizontal table as shown in the figure. $A$ mass of $3 \; kg$ is attached to the free end of the spring. The mass is then pulled sideways to a distance of $2.0 \; cm$ and released. Determine:
$(i)$ the frequency of oscillations,
$(ii)$ maximum acceleration of the mass,and
$(iii)$ the maximum speed of the mass.

$A$ mass $m$ is suspended separately by two different springs,and the time periods are $t_1$ and $t_2$ respectively. If it is connected by both springs in parallel as shown in the figure,then the time period is $t_0$. The correct relation is: