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Question: If U is the PE of an oscillating particle and F is the force acting on it at a given instant. Which ...

If U is the PE of an oscillating particle and F is the force acting on it at a given instant. Which of the following is true?
a. UF+x=0\dfrac{U}{F} + x = 0
b. 2UF+x=0\dfrac{{2U}}{F} + x = 0
c. FU+x=0\dfrac{F}{U} + x = 0
d. F2U+x=0\dfrac{F}{{2U}} + x = 0

Explanation

Solution

The potential energy is defined as the energy possessed by the position of an object. The simple harmonic motion is a motion in which the object oscillates about the mean position and between two extreme points.

Formula used:
The formula of the potential energy is given by,
⇒U=12kx2\Rightarrow U = \dfrac{1}{2}k{x^2}
Where the spring constant is k and the displacement is x.
The formula of the force acting on the oscillating body is given by,
⇒F=−kx\Rightarrow F = - kx
Where force is F, the spring constant is k and the displacement is x.

Complete step by step answer:
Step1:
The Potential energy of a particle executing SHM is given by-
⇒U=12kx2\Rightarrow U = \dfrac{1}{2}k{x^2}………….(1)
Where U= potential energy.
k= spring constant or stiffness.
x= displacement.
Also the force acting on a body executing SHM causing a displacement x is,
⇒F=−kx\Rightarrow F = - kx

Step2:
Now substitute the above in equation (1) and simplifying,
⇒U=12×(kx)×x\Rightarrow U = \dfrac{1}{2} \times \left( {kx} \right) \times x
⇒U=12×(−F)×x\Rightarrow U = \dfrac{1}{2} \times \left( { - F} \right) \times x
⇒2U=−F×x\Rightarrow 2U = - F \times x
⇒2UF=−x\Rightarrow \dfrac{{2U}}{F} = - x
⇒2UF+x=0\Rightarrow \dfrac{{2U}}{F} + x = 0

Hence, the correct answer is option (B).

Additional information:
The simple harmonic motion always has its kinetic energy changing into its potential energy and vice-versa. Whenever the oscillating body is passing through the mean position then its displacement of the body will be zero and at the extremes point the velocity of the body is zero.

Note: The students are advised to understand and remember the formula of the potential energy of the simple harmonic motion and also the formula of the force which is acting on the oscillating body as it is very useful in solving these types of problems. The negative sign in the formula of the force shows that the direction of force is always towards the mean position of the oscillating body.