MHT CET · Physics · Oscillations
The potential energy of 'a long spring when it is stretched by 3 cm is ' \(U\) '. If the spring is stretched by 9 cm , potential energy stored in it will be
- A \(3 \dot{U}\)
- B \(4 U\)
- C 5 U
- D \(9 U\)
Answer & Solution
Correct Answer
(D) \(9 U\)
Step-by-step Solution
Detailed explanation
P.E. stored in spring is, \(\mathrm{U}=\frac{1}{2} \mathrm{kx}^2\)
\(\begin{array}{ll}
\therefore & \frac{U^{\prime}}{U}=\left(\frac{x_2}{x_1}\right)^2=\left(\frac{9}{3}\right)^2=(3)^2=9 \\
\therefore & U^{\prime}=9 U
\end{array}\)
\(\begin{array}{ll}
\therefore & \frac{U^{\prime}}{U}=\left(\frac{x_2}{x_1}\right)^2=\left(\frac{9}{3}\right)^2=(3)^2=9 \\
\therefore & U^{\prime}=9 U
\end{array}\)
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