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Chapter 12: Problem 5

A particle having mass 10 g oscillates according to the equation \(x=(2 \cdot 0 \mathrm{~cm}) \sin \left[\left(100 \mathrm{~s}^{-1}\right) t+\pi / 6\right]\). Find (a) the amplitude, the time period and the spring constant (b) the position, the velocity and the acceleration at \(t=0\).

Short Answer

Expert verified
(a) Amplitude is 2.0 cm, time period is \(\frac{\pi}{50}\) s, spring constant is 100 N/m. (b) At \(t=0\), position is 1.0 cm, velocity is \(100\sqrt{3}\) cm/s, acceleration is -10000 cm/s虏.

Step by step solution

01

Identify Amplitude

The equation of motion is given as \(x = (2.0 \text{ cm}) \sin((100\text{ s}^{-1})t + \frac{\pi}{6})\). The amplitude of the oscillation is the coefficient of the sine function. Thus, the amplitude is \(2.0\text{ cm}\).
02

Find Angular Frequency

Given the equation, the angular frequency \(\omega\) is \(100\text{ s}^{-1}\).
03

Calculate Time Period

The time period \(T\) is related to the angular frequency by the formula \(T = \frac{2\pi}{\omega}\). Substituting \(\omega = 100\text{ s}^{-1}\), we get \(T = \frac{2\pi}{100} = \frac{\pi}{50}\text{ s}\).
04

Determine Spring Constant

The spring constant \(k\) can be found using the formula \(k = m\omega^2\), where \(m = 0.01\text{ kg}\) (mass converted to kg). So, \(k = 0.01 \times (100)^2 = 100\text{ N/m}\).
05

Position at t=0

The position at \(t=0\) is given by substituting \(t=0\) into the equation \(x = (2.0 \text{ cm}) \sin((100\text{ s}^{-1})\cdot 0 + \frac{\pi}{6})\). This simplifies to \(x = 2.0 \times \sin(\frac{\pi}{6}) = 2.0 \times 0.5 = 1.0 \text{ cm}\).
06

Velocity at t=0

The velocity \(v\) is the derivative of \(x\) with respect to \(t\), \(v = \frac{d}{dt}[2.0 \sin((100\text{ s}^{-1})t + \frac{\pi}{6})]\). This leads to \(v = 2.0 \times 100 \times \cos((100\text{ s}^{-1})t + \frac{\pi}{6})\). At \(t=0\), \(v = 200 \times \cos(\frac{\pi}{6}) = 200 \times \frac{\sqrt{3}}{2} = 100\sqrt{3}\text{ cm/s}\).
07

Acceleration at t=0

The acceleration \(a\) is the derivative of velocity with respect to time. Thus, \(a = \frac{d}{dt}[200 \cos((100\text{ s}^{-1})t + \frac{\pi}{6})]\), resulting in \(a = -200 \times 100 \times \sin((100\text{ s}^{-1})t + \frac{\pi}{6})\). At \(t=0\), \(a = -20000 \times \sin(\frac{\pi}{6}) = -20000 \times 0.5 = -10000\text{ cm/s}^2\).

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Key Concepts

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

Amplitude
The amplitude of a simple harmonic motion represents the maximum displacement of the oscillating particle from its equilibrium position. In simpler terms, it鈥檚 the highest point the particle reaches during its motion. For the given problem, the amplitude is extracted directly from the equation:
  • Equation provided: \(x = (2.0 \text{ cm}) \sin((100\text{ s}^{-1})t + \frac{\pi}{6})\)
  • Amplitude: The value \(2.0\text{ cm}\)
This means the particle moves 2 cm away in either direction from the center, reaching a total range of 4 cm. Understanding amplitude is essential because it indicates how "far" the motion goes from its starting position, representing the energy of the oscillation.
Angular Frequency
Angular frequency, denoted as \(\omega\), is a fundamental aspect of simple harmonic motion. It tells us how quickly the oscillation occurs. Think of it as how fast the particle swings back and forth. Angular frequency is tied to the frequency of the motion but expressed in radians per second rather than cycles per second (Hz).
  • From the equation: \((100\text{ s}^{-1})\) is the angular frequency.
This value, \(100 \text{ s}^{-1}\), indicates that the oscillation completes \(100\) radians every second. It's crucial because it influences both the velocity and acceleration of the particle through its direct relationship in these derivatives.
Spring Constant
The spring constant (\(k\)) is a measure of the stiffness of a spring within the context of Hooke鈥檚 Law, connecting directly to the mass and angular frequency of a system in simple harmonic motion. It quantifies the force required to stretch or compress the spring by unit distance. Given the mass of the particle is 0.01 kg, this relation can be established using the formula:
  • \(k = m\omega^2\)
  • With \(m = 0.01\) kg and \(\omega = 100\text{ s}^{-1}\): \(k = 0.01 \times (100)^2 = 100 \text{ N/m}\)
Understanding \(k\) helps predict how the system will behave when disturbed, such as the amount of energy needed to initiate the motion and maintain it against damping forces.
Oscillation
Oscillation describes the repetitive back and forth movement of the particle around an equilibrium position. In simple harmonic motion, this oscillation is sinusoidal, meaning it follows a smooth repetitive wave-like pattern as shown in the motion equation. An oscillation includes movements through a complete cycle, from one dynamic state back to the starting point.
  • The time taken to complete one full cycle is known as the period \(T\).
  • In our scenario, it's calculated using: \(T = \frac{2\pi}{\omega}\) Leading to \(T = \frac{\pi}{50} \text{ seconds}\).
This periodic behavior underlines the prediction of particle positions and velocities over time, essential for understanding waves and vibrations in various physical systems.

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Most popular questions from this chapter

A uniform rod of length \(l\) is suspended by an end and is made to undergo small oscillations. Find the length of the simple pendulum having the time period equal to that of the rod.

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