The total spring constant of the system as shown in the figure will be
$\frac{{{k_1}}}{2} + {k_2}$
${\left[ {\frac{1}{{2{k_1}}} + \frac{1}{{{k_2}}}} \right]^{ - 1}}$
$\frac{1}{{2{k_1}}} + \frac{1}{{{k_2}}}$
${\left[ {\frac{2}{{{k_1}}} + \frac{1}{{{k_2}}}} \right]^{ - 1}}$
In the figure, ${S_1}$ and ${S_2}$ are identical springs. The oscillation frequency of the mass $m$ is $f$. If one spring is removed, the frequency will become
A uniform stick of mass $M$ and length $L$ is pivoted at its centre. Its ends are tied to two springs each of force constant $K$ . In the position shown in figure, the strings are in their natural length. When the stick is displaced through a small angle $\theta $ and released. The stick
Two bodies of masses $1\, kg$ and $4\, kg$ are connected to a vertical spring, as shown in the figure. The smaller mass executes simple harmonic motion of angular frequency $25\, rad/s$, and amplitude $1.6\, cm$ while the bigger mass remains stationary on the ground. The maximum force exerted by the system on the floor is ..... $N$ ( take $g = 10\, ms^{-2}$)
$Assertion :$ The time-period of pendulum, on a satellite orbiting the earth is infinity.
$Reason :$ Time-period of a pendulum is inversely proportional to $\sqrt g$
When a mass $m$ is attached to a spring it oscillates with period $4 \,s$. When an additional mass of $2 \,kg$ is attached to a spring, time period increases by $1 \,s$. The value of $m$ is ........... $kg$