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Question: A particle executing S.H.M has velocities \({V_1}\) and \({V_2}\) at distances \({X_1}\) and \({X_2}...

A particle executing S.H.M has velocities V1{V_1} and V2{V_2} at distances X1{X_1} and X2{X_2} respectively from the mean position. Show that the time period of this particle is T=2πX22−X12V12−V22T = 2\pi \sqrt {\dfrac{{{X_2}^2 - {X_1}^2}}{{{V_1}^2 - {V_2}^2}}} .

Explanation

Solution

In physics, SHM stands for simple harmonic motion. A motion in which a particle oscillates between two fixed points and moves from its one extreme position to another extreme position and hence repeating this motion with a fixed time interval, This time taken by a particle to complete such one oscillation is called Time period of the Simple harmonic motion.

Complete step-by-step solution:
We know that, In a simple harmonic motion, the relation between the velocity of a particle vv at a distance xx from the mean position having angular frequency of ω\omega and of amplitude aa is related as: v2=ω2(a2−x2){v^2} = {\omega ^2}({a^2} - {x^2}) .
Now, in first case we have given that velocity of particle at distance X1{X_1} is V1{V_1} and let aa be the amplitude of this particle’s motion and ω\omega be the angular frequency then we have,
V12=ω2(a2−X12)→(i){V_1}^2 = {\omega ^2}({a^2} - {X_1}^2) \to (i)
And for second case we have given that, velocity of particle at distance X2{X_2} is V2{V_2} so we have,
V22=ω2(a2−X22)→(ii){V_2}^2 = {\omega ^2}({a^2} - {X_2}^2) \to (ii)
Subtract the equation (ii)(ii) from equation (i)(i) we get,
V12−V22=ω2(X22−X12){V_1}^2 - {V_2}^2 = {\omega ^2}({X_2}^2 - {X_1}^2)
Or
V12−V22(X22−X12)=ω2\dfrac{{{V_1}^2 - {V_2}^2}}{{({X_2}^2 - {X_1}^2)}} = {\omega ^2}
Or
T2=(2π)2(X22−X12)V12−V22{T^2} = {(2\pi )^2}\dfrac{{({X_2}^2 - {X_1}^2)}}{{{V_1}^2 - {V_2}^2}}
Or
T=(2π)(X22−X12)V12−V22T = (2\pi )\sqrt {\dfrac{{({X_2}^2 - {X_1}^2)}}{{{V_1}^2 - {V_2}^2}}}
Hence, the time period of the given particle’s Simple Harmonic Motion is T=(2π)(X22−X12)V12−V22T = (2\pi )\sqrt {\dfrac{{({X_2}^2 - {X_1}^2)}}{{{V_1}^2 - {V_2}^2}}} ,proved.

Note: It should be remembered that, The amplitude is the maximum displacement of a particle executing simple harmonic motion from its mean position ( a mean position is one from where the particle starts its motion and an extreme position is one where particle reaches its maximum distance from the mean position) and time period is related with angular frequency as ω=2πT\omega = \dfrac{{2\pi }}{T} . so, the amplitude and the angular frequency remains constant throughout the motion of a particle executing Simple Harmonic Motion.