A particle is excuting a simple harmonic motion. Its maximum acceleration is $\alpha $ and maximum velocity is $\beta $. Then its frequency of vibration will be
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A disc of mass $m$ and radius $R$ is attached to celling with the help of ropes of length $l$. Find the time period of small oscillation of disc in the plane of disc.
The length of a spring is $l$ and its force constant is $k$. When a weight $W$ is suspended from it, its length increases by $x$. If the spring is cut into two equal parts and put in parallel and the same weight $W$ is suspended from them, then the extension will be
Two particles are executing simple harmonic motion of the same amplitude $A$ and frequency $\omega$ along the $x-$axis. Their mean position is separated by distance $X_0(X_0 > A).$ If the maximum separation between them is $(X_0 + A)$, the phase difference between their motion is:
Two pendulum have time periods $T$ and $5T/4$. They start $SHM$ at the same time from the mean position. After how many oscillations of the smaller pendulum they will be again in the same phase
Two particles are in $SHM$ in a straight line. Amplitude $A$ and time period $T$ of both the particles are equal. At time $t=0,$ one particle is at displacement $y_1= +A$ and the other at $y_2= -A/2,$ and they are approaching towards each other. After what time they cross each other ?
The angular frequency of the damped oscillator is given by, $\omega = \sqrt {\left( {\frac{k}{m} - \frac{{{r^2}}}{{4{m^2}}}} \right)} $ where $k$ is the spring constant, $m$ is the mass of the oscillator and $r$ is the damping constant. If the ratio $\frac{{{r^2}}}{{mk}}$ is $8\%$, the change in time period compared to the undamped oscillator is approximately as follows