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Magazine Chapter 15: Problem 23
A particle executes simple harmonic motion with an amplitude of \(3.00 \mathrm{cm} .\) At what position does its speed equal half its maximum speed?
Short Answer
Expert verified
The particle's speed is half its maximum speed at approximately \( \pm 2.60 \mathrm{cm} \) from the equilibrium position.
Step by step solution
01
Understanding Simple Harmonic Motion
A particle in simple harmonic motion (SHM) follows the equation for displacement as a function of time, given by \( x(t) = A\cos(\omega t + \phi) \), where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant. The velocity as a function of time is the derivative of displacement with respect to time, given by \( v(t) = -A\omega\sin(\omega t + \phi) \). The maximum speed \( v_{max} \) is when the sinusoidal function equals 1, i.e., \( v_{max} = A\omega \).
02
Determining Half the Maximum Speed
To find the position where the particle's speed is half its maximum speed, we set \( v = \frac{1}{2}v_{max} \), which gives us \( -A\omega\sin(\omega t + \phi) = \frac{1}{2}A\omega \). The \( A\omega \) terms cancel out, and we are left with \( \sin(\omega t + \phi) = -\frac{1}{2} \).
03
Solving for the Corresponding Displacement
To find the displacement where the speed is half the maximum, we use the fact that \( \sin^2(\omega t + \phi) + \cos^2(\omega t + \phi) = 1 \). Knowing \( \sin(\omega t + \phi) = -\frac{1}{2} \), we can solve for \( \cos(\omega t + \phi) \), which represents the displacement as a fraction of the amplitude. \( \cos(\omega t + \phi) = \sqrt{1 - \sin^2(\omega t + \phi)} = \sqrt{1 - (\frac{1}{2})^2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \). Using the amplitude \( A = 3.00\mathrm{cm} \), the displacement \( x \) is \( x = A\cos(\omega t + \phi) = 3.00\mathrm{cm} \times \frac{\sqrt{3}}{2} \).
04
Calculating the Displacement
Multiplying the amplitude by \( \frac{\sqrt{3}}{2} \), we get \( x = 3.00\mathrm{cm} \times \frac{\sqrt{3}}{2} = 1.5\sqrt{3}\mathrm{cm} \approx 2.60 \mathrm{cm} \). Therefore, the particle's speed is half its maximum speed at approximately \( \pm 2.60 \mathrm{cm} \) from the equilibrium position.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Amplitude in SHM
The amplitude of a simple harmonic oscillator is the maximum extent of its displacement from the equilibrium position. In essence, it is the peak value that the oscillating particle reaches during its motion on either side of the equilibrium point.
When discussing SHM, it's paramount to understand that the amplitude is a constant, determined by the energy possessed by the system and does not change over time unless external forces do work on the system. In our exercise, the amplitude given is 3.00 cm which signifies the farthest distance the particle moves from its rest position.
When discussing SHM, it's paramount to understand that the amplitude is a constant, determined by the energy possessed by the system and does not change over time unless external forces do work on the system. In our exercise, the amplitude given is 3.00 cm which signifies the farthest distance the particle moves from its rest position.
Angular Frequency
Angular frequency, often denoted by the Greek letter omega (\(\omega\)), is a measure of how quickly a particle undergoes one full oscillation. It's intimately connected to the period \(T\) of the motion, which is the time to complete one full cycle.
The relationship between angular frequency and the period is given by \(\omega = \frac{2\pi}{T}\). In SHM, angular frequency tells us the rate of oscillation and is directly proportional to the maximum speed of the oscillator. It is also used when calculating the velocity and displacement at any point during the motion.
The relationship between angular frequency and the period is given by \(\omega = \frac{2\pi}{T}\). In SHM, angular frequency tells us the rate of oscillation and is directly proportional to the maximum speed of the oscillator. It is also used when calculating the velocity and displacement at any point during the motion.
Velocity in SHM
Velocity in SHM is not constant; it varies over time as the particle oscillates. It follows a sinusoidal pattern and is obtained by differentiating the displacement equation with respect to time. The resulting equation for velocity is a cosine function, which is 90 degrees out of phase with the displacement sine function.
This phase difference means that when displacement is at a maximum (amplitude), velocity is zero, and vice versa. At any point in time, the velocity can be found by plugging the time value into the velocity function.
This phase difference means that when displacement is at a maximum (amplitude), velocity is zero, and vice versa. At any point in time, the velocity can be found by plugging the time value into the velocity function.
Maximum Speed of Oscillator
The maximum speed of an oscillator in SHM occurs exactly when it passes through the equilibrium position, where displacement is zero. Because sinusoidal functions oscillate between -1 and +1, the maximum absolute value of velocity, \(v_{max}\), is when \(\sin(\omega t + \phi) = \pm 1\), leading to \(v_{max} = A\omega\), as provided in the exercise.
Understanding this concept helps one to reason that the speed of the particle is a fraction of the maximum speed at other positions in the oscillation, depending on the displacement at the time.
Understanding this concept helps one to reason that the speed of the particle is a fraction of the maximum speed at other positions in the oscillation, depending on the displacement at the time.
Displacement in SHM
Displacement in SHM, indicated as \(x(t)\), tells us the position of the particle at any given time in its oscillatory path. It varies as a cosine function of time, phase shifted by \(\phi\), and modulated by its amplitude, \(A\).
As in the case demonstrated in this exercise, to find the displacement corresponding to a certain condition on speed, we work with trigonometric identities and the known amplitude. We see that displacement directly influences the momentary value of velocity and all other dynamics of the harmonic motion.
As in the case demonstrated in this exercise, to find the displacement corresponding to a certain condition on speed, we work with trigonometric identities and the known amplitude. We see that displacement directly influences the momentary value of velocity and all other dynamics of the harmonic motion.
