A block oscillating on a spring has an amplitude of 20 cm. What will the amplitude be if the total energy is doubled?

A Block Oscillating On A Spring Has An Amplitude Of 20 Cm.

The purpose of this question is to find the amplitude of an oscillating block attached to the spring when energy is doubled.

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The displacement of a particle from its mean position to an extreme position in an oscillating motion possesses some energy. Similarly, in this case, the block at oscillating motion possesses kinetic energy and when it comes to rest, it possesses potential energy. The sum of both kinetic and potential energies gives us the total energy of the oscillating block. 

Expert Answer:

The “to and fro” motion of a body when it is displaced from its mean position is called simple harmonic motion. Energy is conserved in simple harmonic motion due to the continuous movement of the given block from mean to extreme positions. The total mechanical energy of this block will be given as:

\[\text{Total energy (E)}= \text{Kinetic energy (K)} + \text{Potential energy (U)}\]

\[\frac{1}{2}kA^2= \frac{1}{2}mv^2 + \frac{1}{2}kx^2 \]

$k$ is the constant of force which describes that the force is constant with changing motion of the oscillating block. On the other hand, $A$ is the amplitude of this block which describes the covered distance of a block in oscillating motion. The sum of potential and kinetic energy is constant when mechanical energy is conserved during oscillations of a block attached to a spring.

The total mechanical energy of the oscillating block attached to a spring is given by the following formula:

\[\frac{1}{2}kA^2= constant\]

\[E= \frac{1}{2}kA^2\]

To find the amplitude of the oscillating block, we will rearrange the equation as given below:

\[A= \sqrt{\frac{2E}{k}}\]

From the above equation, we conclude that amplitude $A$ is directly proportional to total mechanical energy $E$, which is represented as:

\[A= \sqrt{E}\]

When total mechanical energy $E$ is doubled, the amplitude can be found by taking $A_1$ and $A_2$ at different instances, where $A_2$ is the required amplitude.

\[\frac{A_1}{A_2} = \frac{\sqrt{E}}{\sqrt{2E}}\]

\[\frac{A_1}{A_2}= \frac{1}{\sqrt{2}}\] 

The rearrangement of the above mentioned equation gives us the required equation when energy is doubled:

\[A_2= \sqrt{2}A_1\]

Numerical Result:

\[A_2= \sqrt{2}A_1\]

By putting given value of amplitude represented as  $A_1$ i.e, $A_1$= $20cm$

\[A_2= \sqrt{2}(20)\]

\[A_2= 28.28cm\]

The amplitude will be $28.28cm$ when the total mechanical energy is doubled, and the value of amplitude $A_1$ is $20cm$.


The amplitude of a block oscillating on the spring is  $14cm$. When energy is doubled, what will be the amplitude?

From the above equation, we know that $A$ is directly proportional to $E$.

\[A= \sqrt{E}\]

When E is doubled, the amplitude can be found by taking $A1$ and $A2$ :

\[\frac{A_1}{A_2} = \frac{\sqrt{E}}{\sqrt{2E}}\]

\[\frac{A_1}{A_2}= \frac{1}{\sqrt{2}}\]

\[A_2= \sqrt{2}A_1\]

By putting the given value of amplitude ($A_1$) i.e, $A_1$= $14cm$ 

\[A_2= \sqrt{2}(14)\]

\[A_2= 19.79cm\]

The amplitude will be $19.79cm$ when the $A_1$ is $14cm$ and energy is doubled.

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