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Chapter 15: Problem 2

In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression $$x=(5.00 \mathrm{cm}) \cos (2 t+\pi / 6)$$. where \(x\) is in centimeters and \(t\) is in seconds. At \(t=0\) find (a) the position of the piston, (b) its velocity, and (c) its acceleration. (d) Find the period and amplitude of the motion.

Short Answer

Expert verified
The position of the piston at \(t = 0\) is approximately \(4.33 cm\). The velocity of the piston at \(t = 0\) is approximately \(-5.00 cm/s\). The acceleration of the piston at \(t = 0\) is approximately \(-17.32 cm/s虏\). The period of the motion is \(蟺\) seconds and the amplitude is \(5.00 cm\).

Step by step solution

01

Position of the piston at t=0

At \(t=0\), by substituting the value of \(t\) into the given equation, we get the following: \(x=(5.00 cm) \cos (2 * 0+\pi / 6) = (5.00 cm) \cos(\pi / 6)\). Evaluating the expression for \(x\), we find its value to be approximately \(4.33 cm\).
02

Velocity of the piston at t=0

The velocity function is the derivative of the position function. Therefore, we obtain that by differentiating \(x(t)\) with respect to \(t\). The derivative of \(\cos (2t+\pi / 6)\) is \(-2 \sin (2t+\pi / 6)\). By substituting \(t=0\) in the derivative, we have the velocity of the piston, \(v=-2 * (5.00cm) \sin(\pi / 6)\). After evaluating the expression for \(v\), we determine its value to approximately \(-5.00 cm/s\).
03

Acceleration of the piston at t=0

Acceleration is the derivative of the velocity function. By differentiating \(v(t)\) with respect to \(t\), we have the acceleration \(a = -2 * (5.00 cm) * 2 * \cos (2 * 0 + \pi / 6) = -20.00 cm cos(\pi / 6)\). Evaluating the expression for \(a\), we find its value to be approximately \(-17.32 cm/s虏\).
04

Period and Amplitude of the motion

The period and amplitude of a simple harmonic motion can be directly obtained from the equation of motion. The period \(T\) is given by \(T = 2蟺 / 蠅\), where \(蠅\) is the coefficient of \(t\) in the cosine expression (which is \(2\) in our case) the answer is \(蟺\) (seconds). The amplitude \(A\) is the coefficient of the cosine expression, which, in our case is \(5.00 cm\).

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Key Concepts

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

Oscillation
Oscillation is the repetitive variation or fluctuation between two points about a central value. In the context of simple harmonic motion (SHM), it is the consistent back and forth movement around an equilibrium position. Imagine a piston moving inside an engine; it moves from a point 'A' to point 'B' and back to 'A.' This is a single oscillation. Whether it's a pendulum swinging or the piston in our problem, oscillation is crucial in mechanical systems as it defines the motion pattern. In SHM, this oscillation is regular and predictable, described by mathematical functions like sine or cosine.
Amplitude
The amplitude of an oscillation is the maximum distance the oscillating object moves from its central, or equilibrium, position. It represents the peak of the wave in terms of displacement. In our piston example, the amplitude is given as 5.00 cm. This means that at the peak of its oscillation, the piston is 5.00 cm away from its central position. Amplitude is crucial because it indicates the energy in the system: a larger amplitude means more energy.
Period
The period in simple harmonic motion is the time it takes for the oscillating object to complete one full cycle. This is measured in seconds. For our piston, the motion鈥檚 period can be found using the formula: \[ T = \frac{2\pi}{\omega} \] where \( \omega \) is the angular frequency, which is the coefficient of \( t \) in the position function. Here, \( \omega = 2 \), so the period is \( \pi \) seconds. Understanding the period helps predict when the object will be back in the same position, aiding in synchronizing mechanical systems.
Piston Motion
Piston motion refers to the repetitive back-and-forth movement of a piston within a cylinder, a characteristic of many engines. In simple harmonic motion, piston motion is smooth and predictable. The equation describing the piston's motion in this exercise is \( x = (5.00 \text{ cm}) \cos(2t + \pi/6) \).Here, the piston's position varies periodically over time, demonstrating a clear pattern that highlights the importance of mathematics in engineering and physics. Such predictable behavior is essential for maintaining efficiency in engines.
Position
The position of an object in simple harmonic motion, like our piston, is the location of the object relative to its equilibrium point at a specific time. At any time \( t \), the position \( x \) can be calculated using the equation \( x = (5.00 \text{ cm}) \cos(2t + \pi/6) \).At \( t = 0 \), the position is \( x = (5.00 \text{ cm}) \cos(\pi/6) \approx 4.33 \text{ cm} \).This tells us how far the piston is displaced from its central position at the starting point of the motion.
Velocity
In the context of simple harmonic motion, velocity is the rate of change of position with respect to time. It indicates how fast the piston is moving and in which direction. Mathematically, the velocity is the derivative of the position function.For the piston's motion: \( v(t) = \frac{d}{dt}[x(t)] = -2 \times (5.00 \text{ cm}) \sin(2t + \pi/6) \).At \( t = 0 \), the velocity is calculated as \( v = -2 \times (5.00 \text{ cm}) \sin(\pi/6) \approx -5.00 \text{ cm/s} \).Negative velocity means the piston is moving in the opposite direction to the starting displacement.
Acceleration
Acceleration in simple harmonic motion is the rate of change of velocity over time. It indicates how quickly the object is speeding up or slowing down as it returns to its equilibrium position. The acceleration is found by differentiating the velocity function. For our piston: \( a(t) = \frac{d}{dt}[v(t)] = -2 \times 2 \times (5.00 \text{ cm}) \cos(2t + \pi/6) \).At \( t = 0 \), the acceleration is \( a = -20.00 \text{ cm} \times \cos(\pi/6) \approx -17.32 \text{ cm/s}^2 \).This tells us how rapidly the piston is changing its velocity at that moment in time.

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Most popular questions from this chapter

(a) A hanging spring stretches by \(35.0 \mathrm{cm}\) when an object of mass \(450 \mathrm{g}\) is hung on it at rest. In this situation, we define its position as \(x=0 .\) The object is pulled down an additional \(18.0 \mathrm{cm}\) and released from rest to oscillate without friction. What is its position \(x\) at a time 84.4 s later? (b) What If? A hanging spring stretches by \(35.5 \mathrm{cm}\) when an object of mass \(440 \mathrm{g}\) is hung on it at rest. We define this new position as \(x=0 .\) This object is also pulled down an additional \(18.0 \mathrm{cm}\) and released from rest to oscillate without friction. Find its position 84.4 s later. (c) Why are the answers to (a) and (b) different by such a large percentage when the data are so similar? Does this circumstance reveal a fundamental difficulty in calculating the future? (d) Find the distance traveled by the vibrating object in part (a). (e) Find the distance traveled by the object in part (b).

A \(7.00-\mathrm{kg}\) object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical oscillations having a period of \(2.60 \mathrm{s}\). Find the force constant of the spring.

A \(0.500-\mathrm{kg}\) object attached to a spring with a force constant of \(8.00 \mathrm{N} / \mathrm{m}\) vibrates in simple harmonic motion with an amplitude of \(10.0 \mathrm{cm} .\) Calculate (a) the maximum value of its speed and acceleration, (b) the speed and acceleration when the object is \(6.00 \mathrm{cm}\) from the equilibrium position, and (c) the time interval required for the object to move from \(x=0\) to \(x=8.00 \mathrm{cm}\).

Damping is negligible for a \(0.150-\mathrm{kg}\) object hanging from a light \(6.30-\mathrm{N} / \mathrm{m}\) spring. A sinusoidal force with an amplitude of \(1.70 \mathrm{N}\) drives the system. At what frequency will the force make the object vibrate with an amplitude of \(0.440 \mathrm{m} ?\)

A \(1.00-\mathrm{kg}\) object is attached to a horizontal spring. The spring is initially stretched by \(0.100 \mathrm{m},\) and the object is released from rest there. It proceeds to move without friction. The next time the speed of the object is zero is 0.500 s later. What is the maximum speed of the object?
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