To show that a simple pendulum executes simple harmonic motion, it is necessary to assume that
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(c)If amplitude is large motion will not remain simple harmonic.
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A body executing simple harmonic motion has a maximum acceleration equal to $ 24\,metres/se{c^2} $ and maximum velocity equal to $ 16\;metres/sec $. The amplitude of the simple harmonic motion is
A spring is stretched by $5 \,\mathrm{~cm}$ by a force $10 \,\mathrm{~N}$. The time period of the oscillations when a mass of $2 \,\mathrm{~kg}$ is suspended by it is :(in $s$)
Two pendulums 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 :
A point performs simple harmonic oscillation of period $T$ and the equation of motion is given by $x=Asin$$\left( {\omega t + \frac{\pi }{6}} \right)$. After the elapse of what fraction of the time period the velocity of the point will be equal to half of its maximum velocity?
A particle is performing simple harmonic motion
$(i)$ its velocity-displacement graph is parabolic in nature
$(ii)$ its velocity-time graph is sinusoidal in nature
$(iii)$ its velocity-acceleration graph is elliptical in nature
Correct answer is
A particle of mass $10 \,g$ is undergoing $S.H.M.$ of amplitude $10 \,cm$ and period $0.1 \,s$. The maximum value of force on particle is about ............ $N$