Question:
A particle performing S.H.M. when displacement is 'x', the potential energy and restoring force acting on it are denoted by 'E' and 'F' respectively. The relation between x, E and F is
A particle performing S.H.M. when displacement is 'x', the potential energy and restoring force acting on it are denoted by 'E' and 'F' respectively. The relation between x, E and F is
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You can verify this quickly using signs alone. Restoring force $F$ always acts in the opposite direction of displacement $x$, making their product $F \cdot x$ negative. Since potential energy $E$ is strictly positive, the ratio $\frac{2E}{F}$ must be negative. Adding $x$ (which balances the sign mismatch) is the only way to sum the terms to zero.
Updated On: Jun 12, 2026
- $2EF - x^2 = 0$
- $2EF + x^2 = 0$
- $2EF + x = 0$
- $2EF - x = 0$
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Question:
We need to find the correct algebraic equation linking displacement ($x$), potential energy ($E$), and restoring force ($F$) for a particle undergoing simple harmonic motion (S.H.M.).
Step 2: Key Formula or Approach:
For a particle in simple harmonic motion with a force constant $k$ at a displacement $x$:
1. The restoring force is given by Hooke's Law:
$$F = -kx$$ 2. The potential energy stored in the system is:
$$E = \frac{1}{1}kx^2$$ We can eliminate the system constant $k$ by combining these two baseline equations.
Step 3: Detailed Explanation:
Let's rearrange the potential energy formula to isolate the term $kx^2$:
$$2E = kx^2$$ We can break down $kx^2$ into product components to map it directly to our force term:
$$2E = (kx) \cdot x$$ From the restoring force equation, we know that $kx = -F$. Let's substitute $-F$ back into the broken-down potential energy relation:
$$2E = (-F) \cdot x$$ $$2E = -Fx$$ To find an expression that matches the structures in our options, let's divide both sides by the force $F$:
$$\frac{2E}{F} = -x$$ Move the negative displacement term to the left side of the equation to set it equal to zero:
$$\frac{2E}{F} + x = 0$$ Note: The problem text formats the fraction $\frac{2E}{F}$ inline as $2E/F$. This corresponds precisely to the statement in option (C).
Step 4: Final Answer:
The correct algebraic relationship is $2E/F + x = 0$, which corresponds to option (C).
We need to find the correct algebraic equation linking displacement ($x$), potential energy ($E$), and restoring force ($F$) for a particle undergoing simple harmonic motion (S.H.M.).
Step 2: Key Formula or Approach:
For a particle in simple harmonic motion with a force constant $k$ at a displacement $x$:
1. The restoring force is given by Hooke's Law:
$$F = -kx$$ 2. The potential energy stored in the system is:
$$E = \frac{1}{1}kx^2$$ We can eliminate the system constant $k$ by combining these two baseline equations.
Step 3: Detailed Explanation:
Let's rearrange the potential energy formula to isolate the term $kx^2$:
$$2E = kx^2$$ We can break down $kx^2$ into product components to map it directly to our force term:
$$2E = (kx) \cdot x$$ From the restoring force equation, we know that $kx = -F$. Let's substitute $-F$ back into the broken-down potential energy relation:
$$2E = (-F) \cdot x$$ $$2E = -Fx$$ To find an expression that matches the structures in our options, let's divide both sides by the force $F$:
$$\frac{2E}{F} = -x$$ Move the negative displacement term to the left side of the equation to set it equal to zero:
$$\frac{2E}{F} + x = 0$$ Note: The problem text formats the fraction $\frac{2E}{F}$ inline as $2E/F$. This corresponds precisely to the statement in option (C).
Step 4: Final Answer:
The correct algebraic relationship is $2E/F + x = 0$, which corresponds to option (C).
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