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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
Two pendulums have time periods $T$ and $\frac{{5T}}{4}.$They start $S.H.M.$ at the same time from the mean position. What will be the phase difference between them after the bigger pendulum has complete one oscillation ..... $^o$
The amplitude and the time period in a $S.H.M.$ is $0.5 \,cm$ and $0.4 \,sec$ respectively. If the initial phase is $\pi /2$ radian, then the equation of $S.H.M.$ will be
When a particle of mass $m$ is attached to a vertical spring of spring constant $k$ and released, its motion is described by $y ( t )= y _{0} \sin ^{2} \omega t ,$ where $'y'$ is measured from the lower end of unstretched spring. Then $\omega$ is
A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency $\omega $. The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time
A spring hangs vertically from the ceiling and a mass is attached to its free end. When the mass is pulled down and released, it oscillates vertically with simple harmonic motion of period $T$ . The variation with time $t$ of its distance from the ceiling is as shown. Which statement gives a correct deduction from this graph?
${T}_{0}$ is the time period of a simple pendulum at a place. If the length of the pendulum is reduced to $\frac{1}{16}$ times of its initial value, the modified time
The period of oscillation of a simple pendulum of length $L$ suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination $\alpha$, is given by