STEM Problem Detail

Problem #71

Computer Science Hard

Problem Statement

Prompt

Provide step by step answer of question Example 7.21

Solution Rendering...

**Solution.** We know that the *total energy of a particle executing S.H.M.* $= 2\pi^2 ma^2/T^2$, where the symbols have their usual meanings.

Here, $T = 2\pi$ sec and, therefore, total energy of the particle

$$= \frac{2\pi^2 ma^2}{(2\pi)^2} = \frac{ma^2}{2} = 10240$$

whence, $\dfrac{1}{2} ma^2 = 10240$. Or, $ma^2 = 20480$. $\quad \cdots (i)$

Since the time here is counted from the instant the particle passes through its mean or equilibrium position, we have $\phi = 0$. Hence its *displacement* is given by the relation $x = a \sin \omega t$.

Now, $T = 2\pi/\omega = 2\pi$ (given). So that, $\omega = 1$ radian/sec, $x = 8\sqrt{2}$ cm and $t = \pi/4$.

$$8\sqrt{2} = a \sin\!\left(1 \times \frac{\pi}{4}\right) = a \sin\frac{\pi}{4} = \frac{a}{\sqrt{2}}$$

whence $a = 8\sqrt{2} \times \sqrt{2} = 8 \times 2 = 16$ cm.

*i.e., the amplitude of the particle is 16 cm.*

Substituting this value of $a$ in expression (i) above, we have

$$m \times (16)^2 = 20480 \quad \Rightarrow \quad m = \frac{20480}{256} = 80 \text{ gm.}$$

Thus, *the mass of the particle is 80 gm.*

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