Question 9: Oscillations

Question

A particle executes a simple harmonic motion of amplitude $$\mathrm{A}$$. The distance from the mean position at which its kinetic energy is equal to its potential energy is

Accepted Answer

In simple harmonic motion (SHM), the total mechanical energy of the system is conserved and is given by the sum of kinetic energy (KE) and potential energy (PE). At any point in the motion, the kinetic energy and potential energy can be expressed as:
The total energy (E) in SHM of amplitude $$\mathrm{A}$$ is
$$E = \frac{1}{2} k A^2$$
where $$k$$ is the spring constant.
The potential energy (PE) at a distance $$x$$ from the mean position is given by:
$$\mathrm{PE} = \frac{1}{2} k x^2$$
The kinetic energy (KE) at a distance $$x$$ from the mean position is:
$$\mathrm{KE} = \frac{1}{2} k (A^2 - x^2)$$
We need to find the distance from the mean position where the kinetic energy is equal to the potential energy. Therefore, we set the kinetic energy equal to the potential energy:
$$\frac{1}{2} k x^2 = \frac{1}{2} k (A^2 - x^2)$$
By simplifying the equation, we get:
$$x^2 = A^2 - x^2$$
Adding $$x^2$$ to both sides, we have:
$$2x^2 = A^2$$
Dividing both sides by 2, we get:
$$x^2 = \frac{A^2}{2}$$
Taking the square root of both sides, we find:
$$x = \frac{A}{\sqrt{2}} = \frac{A}{\sqrt{2}} 	imes \frac{\sqrt{2}}{\sqrt{2}} = \frac{A \sqrt{2}}{2} = \frac{\sqrt{2}}{2} A$$
We know that $$\sqrt{2} \approx 1.41$$, so:
$$x \approx \frac{1.41}{2} A \approx 0.71 A$$
Therefore, the distance from the mean position at which the kinetic energy is equal to its potential energy is approximately $$0.71 A$$. Hence, the correct answer is Option B.

Answer

0.91 A

Answer

0.71 A

Answer

0.81 A

Answer

0.51 A

Oscillations

A particle executes a simple harmonic motion of amplitude $\mathrm{A}$. The distance from the mean position at which its kinetic energy is equal to its potential energy is

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