$A$ spring having a spring constant $K$ is loaded with a mass $m$. The spring is cut into two equal parts and one of these is loaded again with the same mass. The new spring constant is

$A$ 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?

Write the expression for the restoring force produced in a spring when a body attached to its end is pulled down by a small displacement $x$.

The motion of a mass on a spring,with spring constant $K$ is as shown in the figure. The equation of motion is given by $x(t) = A \sin \omega t + B \cos \omega t$ with $\omega = \sqrt{\frac{K}{m}}$. Suppose that at time $t = 0$,the position of the mass is $x(0)$ and velocity is $v(0)$,then its displacement can also be represented as $x(t) = C \cos (\omega t - \phi)$,where $C$ and $\phi$ are:

$A$ 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. Let us take the position of the mass when the spring is unstretched as $x = 0$,and the direction from left to right as the positive direction of the $x$-axis. Give $x$ as a function of time $t$ for the oscillating mass if at the moment we start the stopwatch $(t = 0)$,the mass is:
$(a)$ at the mean position,
$(b)$ at the maximum stretched position,and
$(c)$ at the maximum compressed position.
In what way do these functions for $SHM$ differ from each other: in frequency,in amplitude,or in the initial phase?

$A$ particle is suspended from a vertical spring which is executing $S.H.M.$ of frequency $5 \ Hz$. The spring is unstretched at the highest point of oscillation. What is the maximum speed of the particle? (Take $g = 10 \ m/s^2$)