Question

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

$\text{A}\text{. }\dfrac{{{k}_{1}}A}{{{k}_{2}}}$
$\text{B}\text{. }\dfrac{{{k}_{2}}A}{{{k}_{1}}}$
$\text{C}\text{. }\dfrac{{{k}_{1}}A}{{{k}_{1}}+{{k}_{2}}}$
$\text{D}\text{. }\dfrac{{{k}_{2}}A}{{{k}_{1}}+{{k}_{2}}}$

Answer 1

When the block oscillates, the springs will elongate in such a way that the force exerted by both is the same. Use the formula for the spring force and an expression for the extension in the left spring when the block is at the right extreme. This value will be the amplitude of point P.
Formula used:
F=-kx

Complete answer:
As we can see in the given figure, a block of mass M is attached to a combination of two springs. The springs are connected in series. It is said that the block is under simple harmonic motion with an amplitude A.
When the block is at its right extreme, the displacement of the block from the mean position is A. Let the extensions in the springs with spring constants ${{k}_{1}}$ and ${{k}_{2}}$ be ${{x}_{1}}$ and ${{x}_{2}}$ respectively.
Here, ${{x}_{1}}+{{x}_{2}}=A$.
When the block oscillates, the springs will elongate in such a way that the force exerted by both is the same.
That means that $F={{k}_{1}}{{x}_{1}}={{k}_{2}}{{x}_{2}}$.
$\Rightarrow {{x}_{1}}=\dfrac{F}{{{k}_{1}}}$
And
${{x}_{2}}=\dfrac{F}{{{k}_{2}}}$
But ${{x}_{1}}+{{x}_{2}}=A$.
$\Rightarrow \dfrac{F}{{{k}_{1}}}+\dfrac{F}{{{k}_{2}}}=A$
$\Rightarrow F\left( \dfrac{1}{{{k}_{1}}}+\dfrac{1}{{{k}_{2}}} \right)=A$
$\Rightarrow F=\dfrac{A}{\left(\dfrac{1}{{{k}_{1}}}+\dfrac{1}{{{k}_{2}}}\right)}=\dfrac{A{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}$ …. (i).
Since the one end of the left spring is fixed, the amplitude of point P is equal to ${{x}_{1}}$.
And we know that ${{x}_{1}}=\dfrac{F}{{{k}_{1}}}$
Substitute the value of F from (i).
$\Rightarrow {{x}_{1}}=\dfrac{\dfrac{A{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}}{{{k}_{1}}}$
$\Rightarrow {{x}_{1}}=\dfrac{{{k}_{2}}A}{{{k}_{1}}+{{k}_{2}}}$

So, the correct answer is “Option D”.

Note:
When we have a combination of spring, we define the equivalent spring constant of the combination. Let the equivalent spring constant of the combination be ${{k}_{eq}}$.
When the total extension is equal to A, $F={{k}_{eq}}A$
And we found that $\Rightarrow F=\dfrac{A}{\left( \dfrac{1}{{{k}_{1}}}+\dfrac{1}{{{k}_{2}}} \right)}$.
This implies that ${{k}_{eq}}=\dfrac{1}{\left( \dfrac{1}{{{k}_{1}}}+\dfrac{1}{{{k}_{2}}} \right)}$
$\Rightarrow \dfrac{1}{{{k}_{eq}}}=\dfrac{1}{{{k}_{1}}}+\dfrac{1}{{{k}_{2}}}$.