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Two masses $m_1$ and $m_2$ are supended 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
The frequency of oscillations of a mass $m$ connected horizontally by a spring of spring constant $k$ is $4 Hz$. When the spring is replaced by two identical spring as shown in figure. Then the effective frequency is,
The displacement of an object attached to a spring and executing simple harmonic motion is given by $ x= 2 \times 10^{-9}$ $ cos$ $\;\pi t\left( m \right)$ .The time at which the maximum speed first occurs is
A particle starts simple harmonic motion from the mean position. Its amplitude is $a$ and total energy $E$. At one instant its kinetic energy is $3E/4.$ Its displacement at that instant is
A particle of mass $1\, {kg}$ is hanging from a spring of force constant $100\, {Nm}^{-1 .}$ 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 ..... .
When the potential energy of a particle executing simple harmonic motion is one-fourth of its maximum value during the oscillation, the displacement of the particle from the equilibrium position in terms of its amplitude $a$ is
Spring of spring constant $1200\, Nm^{-1}$ is mounted on a smooth frictionless surface and attached to a block of mass $3\, kg$. Block is pulled $2\, cm$ to the right and released. The angular frequency of oscillation is .... $ rad/sec$
$A$ block of mass $M_1$ is hanged by a light spring of force constant $k$ to the top bar of a reverse Uframe of mass $M_2$ on the floor. The block is pooled down from its equilibrium position by $a$ distance $x$ and then released. Find the minimum value of $x$ such that the reverse $U$ -frame will leave the floor momentarily.