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Chapter 3: Problem 63

A weight is attached to a spring and reaches its equilibrium position \((x=0) .\) It is then set in motion resulting in a displacement of $$ x=10 \cos t $$ where \(x\) is measured in centimeters and \(t\) is measured in seconds. See the accompanying figure. a. Find the spring's displacement when \(t=0, t=\pi / 3,\) and \(t=3 \pi / 4\) b. Find the spring's velocity when \(t=0, t=\pi / 3,\) and \(t=3 \pi / 4 .\)

Short Answer

Expert verified
a. 10 cm, 5 cm, -7.07 cm. b. 0 cm/s, -8.66 cm/s, -7.07 cm/s.

Step by step solution

01

Understanding the Displacement Function

The given displacement function is \( x = 10 \cos t \). Here, \( x \) represents the displacement of the spring from its equilibrium position in centimeters, and \( t \) is the time in seconds. The equation describes a simple harmonic motion with an amplitude of 10 cm.
02

Finding Displacement at Specific Times

To find the displacement of the spring at specific times, simply substitute the values of \( t \) into the displacement equation:- When \( t = 0 \): \( x = 10 \cos 0 = 10 \times 1 = 10 \) cm.- When \( t = \pi / 3 \): \( x = 10 \cos(\pi / 3) = 10 \times \frac{1}{2} = 5 \) cm.- When \( t = 3\pi / 4 \): \( x = 10 \cos(3\pi / 4) = 10 \times \left(-\frac{\sqrt{2}}{2}\right) = -5\sqrt{2} \approx -7.07 \) cm.
03

Differentiating the Displacement Function

The velocity of the spring can be found by differentiating the displacement function with respect to time \( t \). The derivative of \( x = 10 \cos t \) is given by:\[ v(t) = \frac{dx}{dt} = -10 \sin t \] where \( v(t) \) is the velocity of the spring.
04

Finding Velocity at Specific Times

Substitute the values of \( t \) into the velocity function \( v(t) = -10 \sin t \):- When \( t = 0 \): \( v = -10 \sin 0 = 0 \) cm/s.- When \( t = \pi / 3 \): \( v = -10 \sin(\pi / 3) = -10 \times \frac{\sqrt{3}}{2} = -5\sqrt{3} \approx -8.66 \) cm/s.- When \( t = 3\pi / 4 \): \( v = -10 \sin(3\pi / 4) = -10 \times \frac{\sqrt{2}}{2} = -5\sqrt{2} \approx -7.07 \) cm/s.

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Key Concepts

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

Displacement Function
In simple harmonic motion, the displacement function represents how far an object moves from its equilibrium position at any given time. In this case, the displacement function is given by \( x = 10 \cos t \). This function tells us the position of a weight attached to a spring as it oscillates around its equilibrium point \((x=0)\). The displacement equation involves:
  • Amplitude (\(10\) cm): This is the maximum distance from the equilibrium position. Here, it means the spring can stretch or compress up to 10 cm away from the center point.
  • Cosine Function (\(\cos t\)): The cosine function indicates that this is a regular oscillation, typical of simple harmonic motion, repeating its cycle over time.
This function allows us to find exact positions of the weight at different times by substituting the desired time \(t\) into the equation. By plugging values into it such as \( t = 0 \), \(\pi/3\), and \(3\pi/4\), we can calculate how far away the weight is from its equilibrium position at those moments.
Velocity Calculation
Finding the velocity of an object in simple harmonic motion involves determining how fast it moves over time, which requires differentiating its displacement function. The velocity is the rate of change of displacement with time. For the given problem, the displacement function is \( x = 10 \cos t \). The velocity can be calculated by taking its first derivative with respect to \( t \):\[ v(t) = \frac{dx}{dt} = -10 \sin t \]Here, \( v(t) \) represents the velocity. The presence of \(-10 \sin t\) reflects:
  • The negative sign shows that as the weight moves away from a maximum point, its velocity actually decreases.
  • The sine function suggests the velocity also varies periodically, like the displacement function; it reaches 0 when the object is at its maximum displacement and peaks when the object passes through the equilibrium position.
By substituting specific times such as \( t = 0\), \(\pi/3\), and \(3\pi/4\) into the velocity equation, we can determine the spring鈥檚 velocity at these moments.
Differentiation
Differentiation is a mathematical technique used to determine the rate of change of a function with respect to one of its variables. In the context of simple harmonic motion, differentiation helps calculate how the displacement of an object changes over time, leading to the discovery of velocity. How it works:
  • Finding Derivatives: To find the velocity from the displacement function \(x = 10 \cos t\), we apply differentiation鈥攆ocusing on the rate at which \(x\) changes with \(t\).
  • Chain Rule Application: Since cosine is a basic trigonometric function, its derivative is \(-\sin \), making the derivative \(-10 \sin t\). This outcome directly follows from the known rule \( \frac{d}{dt}[\cos t] = -\sin t \).
Differentiating turns the problem of finding how fast the spring moves (velocity) into a straightforward calculation from a static position calculation (displacement function). This process is essential for fully understanding not only where the spring is but also how it moves over time.

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