Math / Trigonometry

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

STEP 1

1. The mass $m = 9.00 \, \mathrm{kg}$ .
2. Each rubber band has a length $L = 0.200 \, \mathrm{m}$ .
3. The tension in each rubber band is $T = 3.50 \, \mathrm{N}$ .
4. The tension remains constant during vibration.
5. The force of gravity is ignored.
6. The system undergoes small oscillations, allowing the use of simple harmonic motion principles.

STEP 2

1. Analyze the forces acting on the mass.
2. Determine the effective spring constant for the system.
3. Use the formula for angular frequency in simple harmonic motion.

STEP 3

Analyze the forces acting on the mass:
The tension in each rubber band can be resolved into horizontal and vertical components. Since the system is symmetric, the horizontal components cancel each other out, and the vertical components add up to provide the restoring force.

STEP 4

Determine the effective spring constant for the system:
For small displacements, the restoring force $F$ can be approximated as proportional to the displacement $x$ . The vertical component of the tension in each rubber band is $T \sin(\theta)$ , where $\theta$ is the angle each rubber band makes with the vertical. For small angles, $\sin(\theta) \approx \theta \approx \frac{x}{L}$ .
The total restoring force is $2T \sin(\theta) \approx 2T \frac{x}{L}$ .
Thus, the effective spring constant $k$ is given by:
$k = \frac{2T}{L}$
Substitute the given values:
$k = \frac{2 \times 3.50 \, \mathrm{N}}{0.200 \, \mathrm{m}}$

STEP 5

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:
$\boxed{1.97 \, \mathrm{sec}^{-1}}$

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