Question

The mass M shown in the figure ocillates in simple harmonic motion with amplitude A. The aplitude of the point P is

Answer 1

If we apply force F

Then $F=k_1{x}_1$; $F=k_2{x}_2$

$x_1$ & $x_2$ are displacement of springs

if total distance is x

$\therefore A=x_1+x_2=F(\frac{1}{k_1}+\frac{1}{k_2})$

So $\Rightarrow F=\left(\frac{k_1{k}_2}{k_1+k_2}\right)A$

So Amplitude of the point $x_1=\frac{F}{x_1}=\left(\frac{k_2}{k_1+k_2}\right)A$

$\therefore x_1=\frac{k_2}{k_1+k_2}.A$