QuestionIn the diagram below, a mass $m=9.00 \mathrm{~kg}$ is connected to two rubber bands of length $L=0.200 \mathrm{~m}$ under tension $T=3.50 \mathrm{~N}$. Assuming that the tension remains constant, find the angular frequency of vibration of the mass (in $\mathrm{sec}^{-1}$ ). Ignore the force of gravity and remember the definition of the sine function as the ratio of a triangle's opposite side's length fiver the longest side's length.

Studdy Solution
Use the formula for angular frequency in simple harmonic motion:
The angular frequency $\omega$ is given by:
$$\omega = \sqrt{\frac{k}{m}}$$
Substitute the values for $k$ and $m$:
$$\omega = \sqrt{\frac{\frac{2 \times 3.50}{0.200}}{9.00}}$$
Calculate the angular frequency:
$$k = \frac{7.00}{0.200} = 35.00 \, \mathrm{N/m}$$
$$\omega = \sqrt{\frac{35.00}{9.00}}$$
$$\omega = \sqrt{3.888}$$
$$\omega \approx 1.97 \, \mathrm{sec}^{-1}$$
The angular frequency of vibration of the mass is:
$$\boxed{1.97 \, \mathrm{sec}^{-1}}$$