A particle executes simple harmonic motion. Its amplitude is $8 \,cm$ and time period is $6 \,s$. The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude, is ............. $s$
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A mass $m =100\, gms$ is attached at the end of a light spring which oscillates on a frictionless horizontal table with an amplitude equal to $0.16$ metre and time period equal to $2 \,sec$. Initially the mass is released from rest at $t = 0$ and displacement $x = - 0.16$ metre. The expression for the displacement of the mass at any time $t$ is
A particle is executing simple harmonic motion with a period of $T$ seconds and amplitude a metre. The shortest time it takes to reach a point $\frac{a}{{\sqrt 2 }}\,m$ from its mean position in seconds is
A spring whose unstretched length is $\ell $ has a force constant $k$. The spring is cut into two pieces of unstretched lengths $\ell_1$ and $\ell_2$ where, $\ell_1 = n\ell_2$ and $n$ is an integer. The ratio $k_1/k_2$ of the corresponding force constants, $k_1$ and $k_2$ will be
A mass $m$ is attached to two springs as shown in figure. The spring constants of two springs are $K _1$ and $K _2$. For the frictionless surface, the time period of oscillation of mass $m$ is
The equation of motion of a particle of mass $1\,g$ is $\frac{{{d^2}x}}{{d{t^2}}} + {\pi ^2}x = 0$ where $x$ is displacement (in $m$ ) from mean position. The frequency of oscillation is .... $s$ (in $Hz$ )
A ring of diameter $2m$ oscillates as a compound pendulum about a horizontal axis passing through a point at its rim. It oscillates such that its centre move in a plane which is perpendicular to the plane of the ring. The equivalent length of the simple pendulum is .... $m$