Question

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

• A. $0.08 \text{ m}$
• B. $0.02 \text{ m}$
• C. $0.04 \text{ m}$
• D. $0.06 \text{ m}$

Hint:

Instead of dealing with small decimals early in the calculations, work with ratios! Taking the square root of the simplified ratio directly yields $\frac{x}{A} = \sqrt{\frac{8}{18}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$. Thus, $x = \frac{2}{3} A = \frac{2}{3} (0.06) = 0.04 \text{ m}$ instantly.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem presents an object executing simple harmonic motion. At an unknown position $x$, its kinetic and potential energies are given along with the total motion amplitude. We must solve for this specific displacement distance $x$.

Step 2: Key Formula or Approach:
The total mechanical energy $E$ in simple harmonic motion is a constant conserved value equal to the sum of kinetic energy ($K.E.$) and potential energy ($P.E.$):
$$E = K.E. + P.E.$$ The potential energy and total energy can also be expressed in terms of the force constant $k$:
$$P.E. = \frac{1}{2} k x^2$$ $$E = \frac{1}{2} k A^2$$ Taking the ratio of potential energy to total energy eliminates the unknown stiffness constant $k$:
$$\frac{P.E.}{E} = \frac{\frac{1}{2} k x^2}{\frac{1}{2} k A^2} = \frac{x^2}{A^2}$$

Step 3: Detailed Explanation:
First, find the total energy $E$ by adding the given values:
$$E = 10 \text{ J} + 8 \text{ J} = 18 \text{ J}$$ We are given the amplitude $A = 0.06 \text{ m}$ and $P.E. = 8 \text{ J}$. Set up the energy ratio:
$$\frac{8}{18} = \frac{x^2}{(0.06)^2}$$ Simplify the energy fraction:
$$\frac{4}{9} = \frac{x^2}{0.0036}$$ Isolate $x^2$:
$$x^2 = \frac{4}{9} \times 0.0036 = 4 \times 0.0004 = 0.0016 \text{ m}^2$$ Take the square root of both sides to find $x$:
$$x = \sqrt{0.0016} = 0.04 \text{ m}$$

Step 4: Final Answer:
The distance $x$ from the mean position is $0.04 \text{ m}$, which matches option (C).