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The bob of a simple pendulum of length \(l=40\) in. is released from rest when \(\theta=5^{\circ}\). Assuming simple harmonic motion, determine 1.6 s after release \((a)\) the angle \(\theta,(b)\) the magnitudes of the velocity and acceleration of the bob.

Short Answer

Expert verified
(a) \( \theta \approx 0.085 \) radians; (b) \( |v| \approx 9.95 \text{ in/s} \), \( |a| \approx 32.64 \text{ in/s}^2 \).

Step by step solution

01

Understand SHM Equation for Pendulum

For a simple pendulum, the equation for simple harmonic motion (SHM) is given by \( \theta(t) = \theta_0 \cos \left( \sqrt{\frac{g}{L}} t \right) \), where \( \theta_0 \) is the initial angle, \( g \) is the gravitational acceleration (32.2 ft/s² or 386.4 in/s²), and \( L \) is the length of the pendulum in inches.
02

Convert Angle to Radians

The initial angle \( \theta_0 = 5^{\circ} \) needs to be converted to radians for calculation purposes: \( \theta_0 = \, 5^{\circ} \times \frac{\pi}{180^{\circ}} = \frac{\pi}{36} \approx 0.0873 \) radians.
03

Calculate Angular Frequency \( \omega \)

The angular frequency \( \omega \) is calculated using the formula \( \omega = \sqrt{\frac{g}{L}} \). Substituting \( g = 386.4 \, \text{in/s}^2 \) and \( L = 40 \, \text{in} \), we find \( \omega = \sqrt{\frac{386.4}{40}} \). Calculating this gives \( \omega \approx 3.11 \, \text{rad/s} \).
04

Calculate Angle \( \theta(t) \) at \( t = 1.6 \, s \)

Substitute \( t = 1.6 \, s \), \( \theta_0 = 0.0873 \) radians, and \( \omega \approx 3.11 \, \text{rad/s} \) into the SHM equation: \( \theta(1.6) = 0.0873 \cos(3.11 \times 1.6) \). Calculating this, we determine \( \theta(1.6) \approx 0.085 \, \text{radians} \).
05

Calculate Velocity \( v(t) \) at \( t = 1.6 \, s \)

The velocity of the bob is given by \( v(t) = L \cdot \dot{\theta}(t) = -L \cdot \omega \cdot \theta_0 \sin(\omega t) \). Substitute \( L = 40 \), \( \theta_0 = 0.0873 \), \( \omega \approx 3.11 \), and \( t = 1.6 \) into this equation to find \( v(1.6) = -40 \times 3.11 \times 0.0873 \sin(3.11 \times 1.6) \). Solving this, we find \( v(1.6) \approx 9.95 \text{ in/s} \).
06

Calculate Acceleration \( a(t) \) at \( t = 1.6 \, s \)

The acceleration of the bob is given by \( a(t) = -L \cdot \omega^2 \cdot \theta(t) \). Substitute \( L = 40 \), \( \omega \approx 3.11 \), and \( \theta(t) \approx 0.085 \) into this equation to find \( a(1.6) = -40 \times (3.11)^2 \times 0.085 \). Solving this, we get \( a(1.6) \approx -32.64 \text{ in/s}^2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pendulum Motion
In physics, a pendulum is an object hung from a fixed point that swings back and forth under the influence of gravity. When considering simple harmonic motion, commonly referred to as SHM, it's crucial to understand that the movement can be described using specific mathematical equations.

For a simple pendulum like the one in our exercise, it's released from rest at a small angle. This initial displacement allows the pendulum to swing in a regular back-and-forth pattern, demonstrating the principles of SHM through its predictable motion.
  • The amplitude is the maximum extent of the swing, which starts at 5 degrees in our example.
  • The movement is periodic, meaning it repeats over regular intervals.
  • The pendulum swings are symmetrical on either side of its resting position.
This kind of motion is quintessential because it simplifies the analysis of the pendulum's movement over time, which can be accurately described using trigonometric functions.
Angular Frequency
Angular frequency is a critical concept when analyzing pendulum motion in the context of simple harmonic motion. It relates the speed of oscillation to the forces acting on the pendulum.

Specifically, angular frequency is derived using the relation \( \omega = \sqrt{\frac{g}{L}} \), where \( g \) is the acceleration due to gravity, and \( L \) is the pendulum length. By calculating \( \omega \), we can explore how rapidly the pendulum completes its oscillations.
  • The value of \( \omega \) determines how fast the bob swings back and forth.
  • When \( \omega \) is larger, the pendulum cycles faster.
  • For the pendulum in our exercise, \( \omega \approx 3.11 \, \text{rad/s} \), indicating its oscillatory motion's frequency.
Understanding angular frequency is vital to determining various other characteristics, such as velocity and acceleration, discussed in their respective sections.
Velocity Calculation
Velocity in the realm of pendulum motion is all about how quickly the pendulum's bob moves at any given moment. This isn't just a simple back and forth.Instead, velocity changes as the pendulum swings due to how energy is conserved in motion.

To calculate the velocity, we rely on the differential of the angle with respect to time, given by \( v(t) = -L \cdot \omega \cdot \theta_0 \sin(\omega t) \). This formula helps us see how the initial conditions affect speed:
  • \( L = 40 \) inches is the pendulum's length affecting the velocity's magnitude.
  • \( \theta_0 \) is the initial angle in radians; here ~0.0873 radians.
  • The sine component reflects the dynamic change as the pendulum swings.
For the given exercise, after 1.6 seconds, the velocity of the pendulum bob is roughly 9.95 inches/second, indicating the speed at which it swings through its path.
Acceleration Calculation
Acceleration in the context of a pendulum provides insight into how the bob's speed changes over time. It isn't constant; it varies based on the pendulum's position within its cycle. Calculating acceleration is key to understanding SHM deeply.

The formula used here is \( a(t) = -L \cdot \omega^2 \cdot \theta(t) \), reflecting how acceleration correlates to the angle's shift. Here are its components:
  • The length of the pendulum \( L = 40 \) inches.
  • \( \omega \) is our angular frequency from before.
  • \( \theta(t) \), the angle at a specific time, contributes directly to the force exerted.
In our exercise, at 1.6 seconds into the pendulum's release, the bob's acceleration is approximately \(-32.64 \text{ in/s}^2\). Though initially set into motion gently, this demonstrates how forces like gravity mold the pendulum's motion dynamically throughout its travel.

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