Question

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

• A. $\frac{k_{1}A}{k_2}$
• B. $\frac{k_{2}A}{k_2-k_1}$
• C. $\frac{k_{2}A}{k_1+k_2}$
• D. $\frac{k_{2}A}{k_1}$

Answer 1

Answer: (C)

If a force $F$ is applied to $M$ , say to the right, let A be the distance moved by $M$ . If the system is released, it executes simple harmonic motion of amplitude $A$ . If $A_1$ and $A_2$ ae the extensions in springs $k_1$ and $k_2$ then $A=(A_1+A_2)$ and
$F=k_1{A}_1=k_2{A}_2$
$\Rightarrow A_1=\frac{F}{k_1}$ and $A_2=\frac{F}{k_2}$
$\therefore A=A_1+A_2=F\left(\frac{1}{k_1}+\frac{1}{k_2}\right)$
$=\frac{F\left(k_1+k_2\right)}{k_1{k}_2}$
$\Rightarrow F=\frac{k_1{k}_{2}A}{\left(k_1+k_2\right)}$
The amplitude of point $P=$ amplitude of oscillation of spring $k_1$ which is
$A_1=\frac{F}{k_1}=\frac{k_{2}A}{\left(k_1+k_2\right)}$