Question

The mass $2\text{kg}$ shown in figure oscillates in simple harmonic motion with amplitude $1\text{m}$. The amplitude of point $\text{P}$ is
$(K_1=100\text{N/m},K_2=500\text{N/m})$

• A. $0.2\text{m}$
• B. $5.0\text{m}$
• C. $0.16\text{m}$
• D. $0.83\text{m}$

Answer 1

The correct option is D $0.83\text{m}$

If amplitude of $\text{P}$ is $A_1$ and maximum elongation in spring $2$ is $A_2$ then

$A=A_1+A_2=1\text{m}$

Now, in series combination,

$K_{eq}=\frac{K_1{K}_2}{K_{!}+K_2}$

$F=K_{eq}A=\frac{K_1{K}_2}{K_{!}+K_2}A$

And we know, $F=K_1{A}_1$,

$K_1{A}_1=\frac{K_1{K}_2}{K_{!}+K_2}A$

$\Rightarrow A_1=\frac{K_2}{K_{!}+K_2}A$

$\therefore A=\frac{500}{100+500}\times 1=0.83\text{m}$

Hence, option $\left(\text{D}\right)$ is correct.

Why this question ? Hint: When two springs attached in series, restoring force on both springs is same while when springs are connected in parallel, the elongation is same in both spring.