A particle starts oscillating SHM from its equilibrium position than, the ratio of kinetic energy and potential energy of the particle at time T/12 is a.2:1 b.3:1 c.4:1 d.1:4

Question

Gaurav Gupta · Answer

Khimraj · Answer

Total energy is given as TE = (½)mw 2 A 2 Kinetic energy is given as KE =( ½ )mv 2 = (½)mw 2 A 2 cos 2 (wt). Potential energy is given as PE = TE – KE = (½)mw 2 A 2 sin 2 (wt). Value of wt = (360/T)*(T/12) = 30 So KE/PE = cot 2 (wt) = cot 2 (30) = 3. Hope it clears.

ankit singh · Answer

Let the displacement be x K E = 2 1 ​ K ( A 2 − x 2 ) P E = 2 1 ​ K x 2 Given: K E = P E 2 1 ​ K ( A 2 − x 2 ) = 2 1 ​ K x 2 x = 2 ​ A Let the displacement be x K E = 2 1 ​ K ( A 2 − x 2 ) P E = 2 1 ​ K x 2 Given: K E = P E 2 1 ​ K ( A 2 − x 2 ) = 2 1 ​ K x 2 ​

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A particle starts oscillating SHM from its equilibrium position than, the ratio of kinetic energy and potential energy of the particle at time T/12 is a.2:1 b.3:1 c.4:1 d.1:4

A particle starts oscillating SHM from its equilibrium position than, the ratio of kinetic energy and potential energy of the particle at time T/12 is
a.2:1
b.3:1
c.4:1
d.1:4

3 Answers

Let the displacement be x

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A particle starts oscillating SHM from its equilibrium position than, the ratio of kinetic energy and potential energy of the particle at time T/12 is a.2:1 b.3:1 c.4:1 d.1:4

A particle starts oscillating SHM from its equilibrium position than, the ratio of kinetic energy and potential energy of the particle at time T/12 isa.2:1b.3:1c.4:1d.1:4

3 Answers

Let the displacement be x

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A particle starts oscillating SHM from its equilibrium position than, the ratio of kinetic energy and potential energy of the particle at time T/12 is
a.2:1
b.3:1
c.4:1
d.1:4