Q131 / 160 · TS EAMCET 2022 (18 Jul Shift 1) · Physics

A body starting at $t = 0$ from origin and oscillates simple harmonically with a period of $4 s$ . After what time will its kinetic energy be $75 %$ of its total energy

A
$\frac{1}{2} s$
Correct answer
B
$\frac{1}{3} s$
C
$\frac{1}{4} s$
D
$1 s$

Solution

As we know, $K E = \frac{1}{2} m v^{2} = \frac{1}{2} m ω^{2} (A^{2} - x^{2})$
and $T E = \frac{1}{2} m ω^{2} A^{2}$
As per question,
$K E = \frac{75}{100} T E \Rightarrow \frac{1}{2} m ω^{2} (A^{2} - x^{2}) = \frac{1}{2} m ω^{2} A^{2} \Rightarrow x = \pm \frac{A}{2}$
From mean position, time taken to reach half of the amplitude will be, $\frac{T}{12} = \frac{4}{12} = \frac{1}{3} s$

Hence, the correct answer is option: 'B'.

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A body starting at t=0 from origin and oscillates simple harmonically with a period of 4 s. After what time will its kinetic energy be 75% of its total energy

Solution

Relevant Questions

A body starting at $t = 0$ from origin and oscillates simple harmonically with a period of $4 s$ . After what time will its kinetic energy be $75 %$ of its total energy