# For a questions(Comprehension) of Mechanics refer a photo below.?

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Apr 26, 2018

3, $\frac{a}{2}$
4. $2 \sqrt{a g}$
5. $2 a$

#### Explanation:

3. At the point $B$, the length of the elastic string is $a + a \csc \theta$, and hence the elongation is $a \csc \theta$. So, the force exerted by the string on the bead is $\frac{2 m g}{a} a \csc \theta = 2 m g \csc \theta$

For the net vertical component of the force on the bead to be zero, we have

$2 m g \csc \theta \times \cos \theta = m g \implies$cot theta =1/2#

So, $O B = a \cot \theta = \frac{a}{2}$

4. At the point $C$, the elongation of the spring is $a \sqrt{2}$, so that the elastic potential energy at that point is
$\frac{1}{2} \times \frac{2 m g}{a} \times {\left(a \sqrt{2}\right)}^{2} = 2 m g a$

The gravitational potential energy of the bead at that point is $m g a$ (assuming the reference level of this energy to be at $O$).

At the point $a$, the elastic potential energy is

$\frac{1}{2} \times \frac{2 m g}{a} \times {a}^{2} = m g a$

So, energy conservation gives

$2 m g a + m g a = m g a + \frac{1}{2} m {v}^{2} \implies v = 2 \sqrt{a g}$

5. Let the greatest depth that the bead reaches be $d$. At the lowest point, the elongation of the string is $\sqrt{{a}^{2} + {d}^{2}}$ and so the net potential energy is

$\frac{1}{2} \times \frac{2 m g}{a} \times \left({a}^{2} + {d}^{2}\right) - m g d$

At this point, the bead has to be at rest, so conservation of energy yields

$3 m g a = \frac{m g}{a} \left({a}^{2} + {d}^{2}\right) - m g d \implies {d}^{2} - a d - 2 {a}^{2} = 0 \implies$
$\left(d + a\right) \left(d - 2 a\right) = 0 \implies d = 2 a$