KCET · Physics · Oscillations
The variations of kinetic energy \(\mathrm{K}(\mathrm{x})\), potential energy \(\mathrm{U}(\mathrm{x})\) and total energy as a function of displacement of a particle in SHM is as shown in the figure. The value of \(\left|x_0\right|\) is
- A 2 A
- B \(\frac{\mathrm{A}}{\sqrt{2}}\)
- C \(\sqrt{2} \mathrm{~A}\)
- D \(\frac{A}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{A}}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
\(\text {At } \mathrm{X}_0 \)
\( \mathrm{KE}=\mathrm{PE} \)
\( \frac{1}{2} \mathrm{mw}^2\left(\mathrm{~A}^2-\mathrm{x}^2\right)=\frac{1}{2} \mathrm{mw}^2 \mathrm{x}^2 \)
\( 2 \mathrm{x}^2=\mathrm{A}^2 \)
\( \mathrm{x}=\frac{\mathrm{A}}{\sqrt{2}} \mathrm{x}_0=\frac{\mathrm{A}}{\sqrt{2}}\)
\( \mathrm{KE}=\mathrm{PE} \)
\( \frac{1}{2} \mathrm{mw}^2\left(\mathrm{~A}^2-\mathrm{x}^2\right)=\frac{1}{2} \mathrm{mw}^2 \mathrm{x}^2 \)
\( 2 \mathrm{x}^2=\mathrm{A}^2 \)
\( \mathrm{x}=\frac{\mathrm{A}}{\sqrt{2}} \mathrm{x}_0=\frac{\mathrm{A}}{\sqrt{2}}\)
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