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प्रश्न
A mass attached to a spring is free to oscillate, with angular velocity ω, in a horizontal plane without friction or damping. It is pulled to a distance x0 and pushed towards the centre with a velocity v0 at time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters ω, x0 and v0. [Hint: Start with the equation x = acos (ωt+θ) and note that the initial velocity is negative.]
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उत्तर १
The displacement equation for an oscillating mass is given by:
x = `A cos (omega t + theta)`
Where,
A is the amplitude
x is the displacement
θ is the phase constant
Velocity, `v = (dx)/(dt) = -Aomega sin(omega t + theta)`
At t = 0, x = x0
x0 = Acosθ = x0 … (i)
And, `(dx)/(dt) = -v_0 = Aomegasin theta`
`A sin theta = v_0/omega` ...(ii)
Squaring and adding equations (i) and (ii), we get:
`A^2(cos^2 theta + sin^2 theta)= x_0^2 + ((v_0^2)/(omega^2))`
`:. A= sqrt(x_0^2 + (v_0/omega)^2)`
Hence, the amplitude of the resulting oscillation is `sqrt(x_0^2 + (v_0/omega)^2)`
उत्तर २
x = `alpha cos (omegat + theta)`
`v = (dx)/(dt) = -aomega sin(omegat = theta)`
When t = 0, x = `x_0` and `dx/dt = -v_0`
x0 = `a cos theta`
and `-v_0 = -a omega sin theta` or a `sin theta = v_0/omega`
Squaring and adding (i) and (ii) we get
`a^2(cos^2 theta + sin^2 theta) = x_0^2 + (v_0^2)/(omega^2)`
or `a = sqrt(x_0^2 + v_0^2/omega^2)`
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