To analyze the motion of a body oscillating in simple harmonic motion (SHM) as described by the equation \( x = (6.12 \, \text{m}) \cos[(8.38 \, \text{rad/s})t + 1.92 \, \text{rad}] \), we can break down the problem into several parts. Let's tackle each component step by step.
Finding Displacement
The displacement \( x \) at any time \( t \) is given directly by the equation. To find the displacement at \( t = 1.90 \, \text{s} \), we substitute this value into the equation:
\( x = 6.12 \cos[(8.38 \times 1.90) + 1.92] \)
First, calculate the argument of the cosine function:
- Calculate \( 8.38 \times 1.90 = 15.882 \, \text{rad} \)
- Add \( 1.92 \, \text{rad} \): \( 15.882 + 1.92 = 17.802 \, \text{rad} \)
Now, find \( \cos(17.802) \). Using a calculator, we find:
\( \cos(17.802) \approx 0.073 \)
Therefore, the displacement is:
\( x \approx 6.12 \times 0.073 \approx 0.447 \, \text{m} \)
Calculating Velocity
The velocity \( v \) in SHM can be derived from the displacement function by differentiating it with respect to time \( t \). The formula for velocity is:
\( v(t) = -A \omega \sin(\omega t + \phi) \)
Here, \( A = 6.12 \, \text{m} \) (amplitude), \( \omega = 8.38 \, \text{rad/s} \) (angular frequency), and \( \phi = 1.92 \, \text{rad} \) (phase constant). Substituting \( t = 1.90 \, \text{s} \):
\( v(1.90) = -6.12 \times 8.38 \sin(17.802) \)
We already calculated \( \sin(17.802) \approx -0.997 \). Thus:
\( v(1.90) \approx -6.12 \times 8.38 \times (-0.997) \approx 51.1 \, \text{m/s} \)
Determining Acceleration
Acceleration \( a \) in SHM is found by differentiating the velocity function or using the formula:
\( a(t) = -A \omega^2 \cos(\omega t + \phi) \)
Substituting the known values:
\( a(1.90) = -6.12 \times (8.38)^2 \cos(17.802) \)
Calculate \( (8.38)^2 \approx 70.24 \):
\( a(1.90) \approx -6.12 \times 70.24 \times 0.073 \approx -31.1 \, \text{m/s}^2 \)
Frequency of the Motion
The frequency \( f \) can be calculated from the angular frequency \( \omega \) using the relationship:
\( f = \frac{\omega}{2\pi} \)
Substituting \( \omega = 8.38 \, \text{rad/s} \):
\( f \approx \frac{8.38}{2\pi} \approx 1.33 \, \text{Hz} \)
Period of the Motion
The period \( T \) is the reciprocal of frequency:
\( T = \frac{1}{f} \)
Thus:
\( T \approx \frac{1}{1.33} \approx 0.75 \, \text{s} \)
Summary of Results
- Displacement: \( 0.447 \, \text{m} \)
- Velocity: \( 51.1 \, \text{m/s} \)
- Acceleration: \( -31.1 \, \text{m/s}^2 \)
- Frequency: \( 1.33 \, \text{Hz} \)
- Period: \( 0.75 \, \text{s} \)
By following these steps, we can effectively analyze the motion of the body oscillating in simple harmonic motion. Each calculation builds on the previous one, demonstrating the interconnected nature of these concepts in physics.
