Question

when a 0.50 - kg mass is attached to a vertical spring, the spring stretches by 15 cm. how much mass must be attached to the spring to result in a 0.75 - s period of oscillation?

a. 0.81 kg
b. 0.62 kg
c. 0.28 kg
d. 0.11 kg
e. 0.47 kg

Explanation:

Step1: Find the spring - constant k

Using Hooke's law $F = kx$, where $F=mg$ (the force due to gravity). Given $m = 0.50\ kg$, $x=0.15\ m$ and $g = 9.8\ m/s^{2}$. Since $mg=kx$, then $k=\frac{mg}{x}$. Substituting the values: $k=\frac{0.50\times9.8}{0.15}=\frac{4.9}{0.15}=\frac{490}{15}=\frac{98}{3}\ N/m$.

Step2: Use the formula for the period of a spring - mass system

The formula for the period of a spring - mass system is $T = 2\pi\sqrt{\frac{m}{k}}$. We want to find $m$, and we know $T = 0.75\ s$ and $k=\frac{98}{3}\ N/m$. Rearranging the formula for $m$ gives $m=\frac{kT^{2}}{4\pi^{2}}$.
Substitute $k=\frac{98}{3}$ and $T = 0.75$ into the formula:
\[

$$\begin{align*} m&=\frac{\frac{98}{3}\times(0.75)^{2}}{4\pi^{2}}\\ &=\frac{\frac{98}{3}\times0.5625}{4\pi^{2}}\\ &=\frac{98\times0.5625}{3\times4\pi^{2}}\\ &=\frac{55.125}{3\times4\pi^{2}}\\ &=\frac{55.125}{3\times4\times9.8696}\\ &=\frac{55.125}{118.4352}\\ &\approx0.47\ kg \end{align*}$$

\]

Answer:

e. $0.47\ kg$