In arrangement given in figure, if the block of mass m is displaced, the frequency is given by
$n = \frac{1}{{2\pi }}\sqrt {\left( {\frac{{{k_1} - {k_2}}}{m}} \right)} $
$n = \frac{1}{{2\pi }}\sqrt {\left( {\frac{{{k_1} + {k_2}}}{m}} \right)} $
$n = \frac{1}{{2\pi }}\sqrt {\left( {\frac{m}{{{k_1} + {k_2}}}} \right)} $
$n = \frac{1}{{2\pi }}\sqrt {\left( {\frac{m}{{{k_1} - {k_2}}}} \right)} $
A mass $M$ is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes $S.H.M.$ of time period $T$. If the mass is increased by m, the time period becomes $5T/3$. Then the ratio of $m/M$ is
The spring mass system oscillating horizontally. What will be the effect on the time period if the spring is made to oscillate vertically ?
To make the frequency double of a spring oscillator, we have to
Two masses $M_{A}$ and $M_{B}$ are hung from two strings of length $l_{A}$ and $l_{B}$ respectively. They are executing SHM with frequency relation $f_{A}=2 f_{B}$, then relation
A particle at the end of a spring executes simple harmonic motion with a period ${t_1}$, while the corresponding period for another spring is ${t_2}$. If the period of oscillation with the two springs in series is $T$, then