Question
A spring stretches by 2 mm when it is loaded with a mass of 200 g . From equilibrium position the mass is further pulled down by 2 mm and released. The frequency associated with the system and maximum energy in the spring are $\_\_\_\_$ Hz and
$\_\_\_\_$ J , respectively. (Take $g = 10\text{ }m/s^{2}$ )
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Solution
Sol. [Key Concept: SHM frequency $f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$ with $k$ from $kx = mg$; maximum energy $E = \frac{1}{2}\ k\ x_{\max}^{2}$.]
$kx = 0.2\text{ }g$;
$$f = \frac{1}{2\pi}\sqrt{\frac{k}{\text{ }m}}$$
$${k \times 2 \times 10^{- 3} = 2f = \frac{1}{2\pi}\sqrt{\frac{1000}{0.2}} }{k = 1000\text{ }N/m}$$
$$= \frac{1}{2\pi}\sqrt{5000} = \frac{5}{\pi}\sqrt{50}$$
$$E_{\text{max}\text{~}} = \frac{1}{2}k\ x_{\text{max}\text{~}}^{2} $$$\frac{1}{2} \times 1000 \times \left( 4 \times 10^{- 3} \right)^{2} = 8 \times 10^{- 3}$ Joule
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