A particle is executing simple harmonic motion of amplitude r. At what distance from the mean position, its kinetic energy is equal to its potential energy or the potential energy is half of the kinetic energy.
Ans: \(y=\pm\frac{r}{\sqrt{2}}=\pm0.71r\).
The total energy of a particle executing S.H.M. of period \(2\pi Sec\) is \(10240erg\). The displacement of the particle at \(\frac{\pi}{4}Sec \) is \(8\sqrt{2}cm\). calculate: the amplitude of the motion and mass of the particle.
Ans: \(r=16cm\) and \(m=80g\).
At any instant of time the displacement of a particle executing S.H.M. is half of its amplitude. What fraction of the total energy is kinetic energy and what fraction is potential energy?
Ans: \(K.E.=\frac{3}{4}E\) and \(P.E.=\frac{1}{4}E\).
A point particle of mass \(200g\) is executing S.H.M. with the amplitude of \(0.2m\). When the particle passes through the mean position its kinetic energy is \(\left(16\times10^{-3}\right)J\). Obtain the equation of motion of this particle executing simple harmonic motion if the initial phase of the oscillation is \(4^{o}\).
Ans: \(y=0.2Sin\left(2t+\frac{\pi}{4}\right)\).
A mass of \(2kg\) is executing simple harmonic motion which is given by \(x=10.0Cos\left(100t+\frac{\pi}{4}\right)\) cm. What is the maximum kinetic energy?
Ans: \(K.E_{max}=100J\).
A particle of mass \(100g\) is describing S.H.M. along a straight line with a period of \(2s\) and amplitude of \(12cm\). What is the kinetic energy of the particle when it is (i) \(3cm\) (ii) \(5cm\) from its equilibrium position? Use: \(\pi^{2}=10\).
Ans:(i) \(K.E_{x=3cm}=67500erg\) and (ii) \(K.E_{x=5cm}=59500erg\).
calculate: the ratio of the displacement to amplitude when K.E of a particle executing S.H.M. is thrice the P.E.
