Question

A block is released from point $\text{A}$ as shown in figure. All surfaces are smooth and there is no loss of mechanical energy anywhere. Find the time period of oscillations of block.

• A. $2\sqrt{\frac{2h}{g}}[\frac{1}{\sin \alpha }+\frac{1}{\sin \beta }]$
• B. $2\sqrt{\frac{h}{g}}[\frac{1}{\sin \alpha }+\frac{1}{\sin \beta }]$
• C. $2\sqrt{\frac{2h}{g}}[\sin \alpha +\cos \beta ]$
• D. $2\sqrt{\frac{2h}{g}}[\frac{1}{\sin \alpha }+\sin \beta ]$

Answer 1

The correct option is A $2\sqrt{\frac{2h}{g}}[\frac{1}{\sin \alpha }+\frac{1}{\sin \beta }]$


As there is no loss of mechanical energy, the block rises upto the same height $h$ on other side also,

From figure, $\text{AB}=\frac{h}{\sin \alpha },\text{BC}=\frac{h}{\sin \beta }$

Acceleration of block on the left inclined plane, $a_1=g\sin \alpha$

Acceleration of block on the right inclined plane, $a_2=g\sin \beta$

Velocity at points $\text{A}$ and $\text{C}$ will be zero .i.e.,$u_A=u_C=0$

Using $t=\sqrt{\frac{2s}{a}}$ with $(u=0)$

Time period of oscillations,

$T=2t_{AB}+2t_{CB}=2(t_{AB}+t_{CB})$

Now, substituting the values, in the above equation,

$T=2[\sqrt{\frac{2\times AB}{a_1}}+\sqrt{\frac{2\times CD}{a_2}}]$

$\Rightarrow T=2[\sqrt{\frac{2\times \frac{h}{\sin \alpha }}{g\sin \alpha }}+\sqrt{\frac{2\times \frac{h}{\sin \beta }}{g\sin \beta }}]$

$\therefore T=2\sqrt{\frac{2h}{g}}[\frac{1}{\sin \alpha }+\frac{1}{\sin \beta }]$

Hence, option (a) is correct answer.

Why this question ? The question demands the basic understanding of oscillatory motion and how to calculate time period.