Question 332: Oscillations

Question

An object executes SHM along $$x$$-axis with amplitude $$0.06 \mathrm{~m}$$. At certain distance '$$\mathrm{x}$$' metre from mean position, it has kinetic energy $$10 \mathrm{~J}$$ and potential energy $$8 \mathrm{~J}$$. the distance '$$\mathrm{x}$$' will be

Accepted Answer

In Simple Harmonic Motion (SHM), the total energy (E) of the system remains constant and is the sum of its kinetic energy (K.E) and potential energy (P.E). Therefore, we have:
$$E = 	ext{K.E} + 	ext{P.E}$$
Given that the kinetic energy is $$10 \mathrm{~J}$$ and the potential energy is $$8 \mathrm{~J}$$, the total energy can be calculated as:
$$E = 10 \mathrm{~J} + 8 \mathrm{~J} = 18 \mathrm{~J}$$
The total energy in SHM can also be expressed in terms of amplitude (A) and the maximum potential energy which occurs at the extreme position. Hence, the total energy is given by:
$$E = \frac{1}{2} k A^2$$
where $ k $ is the force constant. Since the amplitude (A) is $$0.06 \mathrm{~m}$$, we can substitute the values to get:
$$18 \mathrm{~J} = \frac{1}{2} k (0.06)^2$$
Solving for $ k $, we get:
$$k = \frac{2 	imes 18}{(0.06)^2}$$
$$k = \frac{36}{0.0036}$$
$$k = 10000 \mathrm{~N/m}$$
Now, using the given kinetic and potential energy values at a distance $ x $ from the mean position, we know that:
$$	ext{K.E} = \frac{1}{2} k (A^2 - x^2)$$
$$	ext{P.E} = \frac{1}{2} k x^2$$
Given that:
$$10 \mathrm{~J} = \frac{1}{2} k (A^2 - x^2)$$
$$8 \mathrm{~J} = \frac{1}{2} k x^2$$
First, let's solve for $ x $ from the potential energy equation:
$$8 = \frac{1}{2} k x^2$$
$$x^2 = \frac{16}{k}$$
Substituting $ k = 10000 \mathrm{~N/m} $:
$$x^2 = \frac{16}{10000}$$
$$x^2 = 0.0016 \mathrm{~m}^2$$
$$x = 0.04 \mathrm{~m}$$
Therefore, the distance '$$x$$' from the mean position where the kinetic energy is $$10 \mathrm{~J}$$ and the potential energy is $$8 \mathrm{~J}$$ is:
Option C: 0.04 m

Answer

0.08 m

Answer

0.02 m

Answer

0.04 m

Answer

0.06 m

Oscillations

An object executes SHM along $x-axis with amplitude 0.06 \mathrm{~m}. At certain distance '\mathrm{x}' metre from mean position, it has kinetic energy 10 \mathrm{~J} and potential energy 8 \mathrm{~J}. the distance '\mathrm{x}$' will be

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