If the distance $x$ of a point on a straight line measured from a fixed origin on it and its velocity $v$ are connected by the relation $4 v^{2} = 25 - x^{2}$ . Find the time period of oscillation in seconds.

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Solution

$4 v^{2} = 25 - x^{2}$
Differentiating,
$v \frac{d v}{d x} \times 8 = - 2 x$
or acceleration $= \frac{d v}{d t} = \frac{d v}{d x} \frac{d x}{d t} = v \frac{d v}{d x} = \frac{- x}{4}$
So, acceleration is proportional to displacement and directed towards the fixed point. So, it executes SHM.
$∴ \frac{d^{2} x}{d t^{2}} = \frac{- x}{4}$
Comparing it with $\frac{d^{2} x}{d t^{2}} = - ω^{2} x \Rightarrow ω^{2} = \frac{1}{4}$
$∴ ω = \frac{1}{2}$
$∴ T =$ period $= \frac{2 π}{ω} = \frac{2 π}{1 / 2} = 4 π sec$

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If the distance x of a point on a straight line measured from a fixed origin on it and its velocity v are connected by the relation 4v2=25−x2. Find the time period of oscillation in seconds.

4v2=25−x2 Differentiating, vdvdx×8=−2x or acceleration =dvdt=dvdxdxdt=vdvdx=−x4 So, acceleration is proportional to displacement and directed towards the fixed point. So, it executes SHM. ∴d2xdt2=−x4 Comparing it with d2xdt2=−ω2x⇒ω2=14 ∴ω=12 ∴T= period =2πω=2π1/2=4π sec

If the distance $x$ of a point on a straight line measured from a fixed origin on it and its velocity $v$ are connected by the relation $4 v^{2} = 25 - x^{2}$ . Find the time period of oscillation in seconds.