A particle of mass 1 kg is hanging from a spring of force constant 100 Nm-1. The mass is pulled slightly downward and released so that it executes free simple harmonic motion with time period T. The time when the kinetic energy and potential energy of the system will become equal, is
The value of x is________.
A particle of mass 1 kg is hanging from a spring of force constant 100 Nm-1. The mass is pulled slightly downward and released so that it executes free simple harmonic motion with time period T. The time when the kinetic energy and potential energy of the system will become equal, is The value of x is________.
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1 Answer
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(Eq. of a particle executing SHM)
When KE = PE
A2 – x2 = x2
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Then,
Given mg = kL
∴ Iα = (kLθ.L + k (L/2)²θ - mg (L/2)θ)
(mL²/3)α = kL² (3/4)θ (restoring torque)
α = (9k/4m)θ
∴ ω = (3/2)√ (k/m)
y = A sin (2πt/T)
t? - t? = (T/2π) [sin? ¹ (x? /A) - sin? ¹ (x? /A)]
Displacement equation of SHM of frequency ' '
Now,
Potential energy
So frequency of potential energy
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