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: List the data.
An SHM has amplitude $A = 0.06\ \text{m}$. At some position $x$, kinetic energy $= 10\ \text{J}$ and potential energy $= 8\ \text{J}$. Find $x$.
Step 2: Get the total energy.
Energy is conserved, so $E = K + U = 10 + 8 = 18\ \text{J}$. This is the same as the energy at the extreme position.
Step 3: Recall the potential energy form.
In SHM $U = \tfrac{1}{2} k x^2$ and the total $E = \tfrac{1}{2} k A^2$.
Step 4: Take a ratio to remove $k$.
$\dfrac{U}{E} = \dfrac{x^2}{A^2}$, which is clean because $k$ cancels.
Step 5: Plug in numbers.
$\dfrac{8}{18} = \dfrac{x^2}{(0.06)^2}$. Simplify $\dfrac{8}{18} = \dfrac{4}{9}$, so $x^2 = \dfrac{4}{9} \times 0.0036 = 0.0016\ \text{m}^2$.
Step 6: Solve for $x$.
$x = \sqrt{0.0016} = 0.04\ \text{m}$, option (3). As a check, $\dfrac{x}{A} = \dfrac{0.04}{0.06} = \dfrac{2}{3}$, and $(2/3)^2 = 4/9$ matches the energy fraction.
\[ \boxed{x = 0.04\ \text{m}} \]