A particle executes simple hormonic motion between $x =\, -A$ and $x = +A$ . It starts from $x = 0$ moves in $+x-$ direction. The time taken for it to move from $x = 0$ to $x = \frac {A}{2}$ is $T_1$ and to move from $\frac {A}{2}$ to $\frac {A}{\sqrt 2}$ is $T_2$ , then
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Equations ${y_1} = A\sin \omega t$ and ${y_2} = \frac{A}{2}\sin \omega t + \frac{A}{2}\cos \omega t$ represent $S.H.M.$ The ratio of the amplitudes of the two motions is
The maximum speed of a particle executing $S.H.M.$ is $1\,m/s$ and its maximum acceleration is $1.57\,m/se{c^2}$. The time period of the particle will be .... $\sec$
A lift is descending with acceleration $g/3$ . What will be the time period of a simple pendulum suspended from its ceiling if its time period in staionary life is $'T'$ ?
A particle is executing $SHM$ with amplitude $A,$ time period $T,$ maximum acceleration $a_o$ and maximum velocity $v_0.$ Its starts from mean position at $t=0$ and at time $t,$ it has the displacement $A/2,$ acceleration $a$ and velocity $v$ then
A particle executes $SHM.$ Its velocities are $v_1$and $v_2$ at displacement $x_1$ and $x_2$ from mean position respectively. The frequency of oscillation will be
The energy of a particle executing simple harmonic motion is given by $E = Ax^2 + Bv^2$, where $x$ is the displacement from mean position $x = 0$ and $v$ is the velocity of the particle at $x$ then choose the incorrect statement