A small ball of mass m is attached to the middle of a tightly stretched perfectly flexible wire AB of length 2 l (figure). The ball is given a small lateral displacement in horizontal direction and released. The initial tension (T) in the wire is high and change in it due to small lateral displacement of the ball can be neglected. Prove that the ball will perform simple harmonic motion, and calculate the period. If there is a device which can change the tension in the wire at will , how will the time period change if tension in the wire is increased?

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$π \sqrt{\frac{mℓ}{2 T}}$

$2 π \sqrt{\frac{mℓ}{T}}$

$π \sqrt{\frac{mℓ}{T}}$

$2 π \sqrt{\frac{mℓ}{2 T}}$

answer is D.

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Detailed Solution

When the ball has a small displacement as shown in the figure, the wires will be inclined slightly to the vertical.
The horizontal component of tension will be the net unbalanced force on the ball
$\begin{array}{l} ma = - 2 T \frac{x}{ℓ} \\ Or, a = - \frac{2 T}{m} \frac{1}{ℓ} x \end{array}$
This is the equation of simple harmonic motion where
$ω^{2} = \frac{2 T}{ml}$
Time period is
$t = \frac{2 π}{ω} = 2 π \sqrt{\frac{mℓ}{2 T}}$
We see that the time period is inversely proportional to the square root of the tension. Thus by increasing the tension the time period of oscillation will decrease.