$A$ 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

$A$ small mass $m$, attached to one end of a spring with a negligible mass and an unstretched length $L$, executes vertical oscillations with angular frequency $\omega_{0}$. When the mass is rotated with an angular speed $\omega$ by holding the other end of the spring at a fixed point, the mass moves uniformly in a circular path in a horizontal plane. Then the increase in length of the spring during the rotation is

Which type of spring has fast oscillations: stiff or soft?

$A$ block of mass $m$ is suspended separately by two different springs having time periods $t_1$ and $t_2$. If the same mass is connected to a parallel combination of both springs,then its time period will be

How does the period of oscillation depend on the mass of the block attached to the end of a spring?

Two blocks of masses $m$ and $M$ $(M > m)$ are placed on a frictionless table as shown in the figure. $A$ massless spring with spring constant $k$ is attached to the lower block. If the system is slightly displaced and released,then ($\mu =$ coefficient of friction between the two blocks):
$(A)$ The time period of small oscillation of the two blocks is $T = 2\pi \sqrt{\frac{M + m}{k}}$
$(B)$ The acceleration of the blocks is $a = \frac{kx}{M + m}$ ($x =$ displacement of the blocks from the mean position)
$(C)$ The magnitude of the frictional force on the upper block is $f = \frac{mkx}{M + m}$
$(D)$ The maximum amplitude of the upper block,if it does not slip,is $A = \frac{\mu mg(M + m)}{mk} = \frac{\mu g(M + m)}{k}$ (Wait,let's re-evaluate: $f_{max} = \mu mg$. Since $f = ma = m \cdot \frac{kx}{M+m}$,at max amplitude $A$,$m \cdot \frac{kA}{M+m} = \mu mg \implies A = \frac{\mu g(M+m)}{k}$)
$(E)$ Maximum frictional force can be $\mu mg$.
Choose the correct answer from the options given below.