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A particle is executing simple harmonic motion with angular frequency ω, then the angular frequency of its kinetic energy will be

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Important Questions on Simple Harmonic Motion

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In a simple harmonic oscillator, at the mean position
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A body executes simple harmonic motion. The potential energy (PE), kinetic energy (KE) and total energy (TE) are measured as a function of displacement (x). Which of the following statements is true?

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The particle executing simple harmonic motion has a kinetic energy K0cos2ωt. The maximum values of the potential energy and the total energy are respectively (Assuming potential energy to be zero at the mean position)
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The total energy of a particle, executing simple harmonic motion is

(Where x is the displacement from the mean position.)
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A particle of mass m executes simple harmonic motion with amplitude a and frequency v. The average kinetic energy during its motion from the position of equilibrium to the end is

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A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy is Ux=kx3 , where k is a positive constant. If the amplitude of oscillation is a, then its time period T is
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Two bodies M and N of equal masses are suspended from two separate massless springs of spring constants k1 and k2, respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of vibration of M to that on N is
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A mass m is suspended by means of two coiled springs which have the same length in unstretched condition as in figure. Their force constants are k1 and k2, respectively. When set into vertical vibrations, the period will be,

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