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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

Kavan , 7 Years ago
Grade 11
anser 3 Answers
Gaurav Gupta

Last Activity: 7 Years ago

To determine the ratio of kinetic energy (KE) to potential energy (PE) of a particle oscillating in simple harmonic motion (SHM) at a specific time, we can use the properties of SHM and the formulas for kinetic and potential energy. In SHM, the total mechanical energy is constant and is the sum of kinetic and potential energy, which can be expressed in terms of amplitude and angular frequency.

Understanding the Motion of SHM

In SHM, a particle oscillates back and forth around an equilibrium position. The displacement of the particle from equilibrium can be described by the equation:

x(t) = A sin(ωt)

Where:

  • A = amplitude of the oscillation
  • ω = angular frequency (ω = 2π/T, where T is the period)
  • t = time

Energy in SHM

The total mechanical energy (E) in SHM is given by:

E = KE + PE

Where:

  • KE = (1/2)mω²(A² - x²)
  • PE = (1/2)mω²x²

At any point in the motion, the sum of KE and PE equals the total energy E, which remains constant.

Calculating Energies at Time T/12

At time T/12, we substitute t into the displacement equation:

x(T/12) = A sin(ω(T/12)) = A sin(π/6) = A(1/2)

Now, we can find the kinetic and potential energies:

Potential Energy

Using the displacement x = A/2, we calculate the potential energy:

PE = (1/2)mω²(A/2)² = (1/2)mω²(A²/4) = (1/8)mω²A²

Kinetic Energy

To find the kinetic energy, we use the total energy, which is constant. The total energy E is:

E = (1/2)mω²A²

Now we can express kinetic energy:

KE = E - PE = (1/2)mω²A² - (1/8)mω²A² = (4/8)mω²A² - (1/8)mω²A² = (3/8)mω²A²

Finding the Ratio of KE to PE

Now that we have both KE and PE, we can find their ratio:

KE/PE = ((3/8)mω²A²) / ((1/8)mω²A²) = 3/1

Conclusion

The ratio of kinetic energy to potential energy of the particle at time T/12 is thus 3:1. Therefore, the correct answer is b. 3:1.

Khimraj

Last Activity: 7 Years ago

Total energy is given as TE = (½)mw2A2
Kinetic energy is given as KE =( ½ )mv2 =  (½)mw2A2cos2(wt).
Potential energy is given as PE = TE – KE = (½)mw2A2sin2(wt).
Value of wt = (360/T)*(T/12) = 30
So KE/PE = cot2(wt) = cot2(30) = 3.
Hope it clears.

ankit singh

Last Activity: 4 Years ago

Let the displacement be x

KE=21K(A2x2)
PE=21Kx2
Given:
KE=PE
21K(A2x2)=21Kx2
x=2ALet the displacement be x
KE=21K(A2x2)
PE=21Kx2
Given:
KE=PE
21K(A2x2)=21Kx2
 

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