/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 61 A weight is attached to a spring... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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
At \(t=0\), displacement is 10 cm and velocity is 0 cm/s. At \(t=\pi/3\), displacement is 5 cm, velocity is -8.66 cm/s. At \(t=3\pi/4\), displacement is -7.07 cm, velocity is -7.07 cm/s.

Step by step solution

01

Determine Displacement Formula

The displacement is given by the function \(x = 10 \cos t\). To find the displacement at specific times, we need to substitute the given \(t\) values into this formula.
02

Calculate Displacement at t=0

Substitute \(t=0\) into the displacement formula: \(x = 10 \cos(0) = 10 \times 1 = 10\). So, the displacement at \(t=0\) is 10 cm.
03

Calculate Displacement at t=\pi / 3

Substitute \(t = \pi / 3\) into the displacement formula: \(x = 10 \cos(\pi / 3) = 10 \times \frac{1}{2} = 5\). So, the displacement at \(t = \pi / 3\) is 5 cm.
04

Calculate Displacement at t=3\pi / 4

Substitute \(t = 3\pi / 4\) into the displacement formula: \(x = 10 \cos(3\pi / 4) = 10 \times \left(-\frac{\sqrt{2}}{2}\right) \approx -7.07\). So, the displacement at \(t = 3\pi / 4\) is approximately -7.07 cm.
05

Determine Velocity Formula

The velocity is the derivative of the displacement function with respect to time, \(v(t) = \frac{dx}{dt}\). Differentiating \(x = 10 \cos t\) gives \(v(t) = -10 \sin t\).
06

Calculate Velocity at t=0

Substitute \(t=0\) into the velocity formula: \(v = -10 \sin(0) = 0\). So, the velocity at \(t=0\) is 0 cm/s.
07

Calculate Velocity at t=\pi / 3

Substitute \(t = \pi / 3\) into the velocity formula: \(v = -10 \sin(\pi / 3) = -10 \times \frac{\sqrt{3}}{2} = -5\sqrt{3} \approx -8.66\). So, the velocity at \(t = \pi / 3\) is approximately -8.66 cm/s.
08

Calculate Velocity at t=3\pi / 4

Substitute \(t = 3\pi / 4\) into the velocity formula: \(v = -10 \sin(3\pi / 4) = -10 \times \frac{\sqrt{2}}{2} \approx -7.07\). So, the velocity at \(t = 3\pi / 4\) is approximately -7.07 cm/s.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Spring Displacement Explained
Springs are fascinating mechanical devices that can store and release energy through their motion. When a weight is attached to a spring and is set in motion, the spring stretches and compresses, creating a back-and-forth motion. This is a type of harmonic motion, and it can be described mathematically using a displacement function. The function provided in the exercise is \[x = 10 \cos t,\]where \(x\) is the displacement measured in centimeters, and \(t\) is the time in seconds.

To find out how far the spring has moved from its equilibrium point at different times, substitute the given time values into the displacement equation. The cosine function, \(\cos t\), tells us how much the spring is displaced from the neutral position, multiplying it by 10 gives the displacement in centimeters. For instance:
  • At \(t = 0\), \(x = 10 \times \cos(0) = 10\) cm. The spring is stretched to its maximum position.
  • At \(t = \pi/3\), \(x = 10 \times \cos(\pi/3) = 5\) cm. The displacement is half the amplitude.
  • At \(t = 3\pi/4\), \(x = 10 \times \cos(3\pi/4) \approx -7.07\) cm. The negative sign indicates the spring is on the opposite side of the equilibrium position.
This illustrates how the spring oscillates over time, with a maximum displacement of 10 cm.
Velocity Calculation in Harmonic Motion
Velocity is crucial in understanding how fast an object is moving and in which direction. In the context of harmonic motion for a spring, velocity comes from the derivative of the displacement function. Using the given formula \( x = 10 \cos t \), the velocity \( v(t) \) can be derived as \[v(t) = \frac{dx}{dt} = -10 \sin t.\]This shows that velocity is proportional to the sine of time and indicates how the speed of the spring's movement varies.

With this formula, you can calculate the velocity at specific times. For example:
  • At \(t = 0\), \(v = -10 \sin(0) = 0\) cm/s. The spring is momentarily stationary as it changes direction.
  • At \(t = \pi/3\), \(v = -10 \times \sin(\pi/3) \approx -8.66\) cm/s. This suggests the spring is moving fast in the opposite direction.
  • At \(t = 3\pi/4\), \(v = -10 \times \sin(3\pi/4) \approx -7.07\) cm/s. The spring continues its motion, maintaining significant speed.
Understanding velocity helps predict the motion dynamics of the spring, showing when it speeds up and slows down through its oscillation.
Role of Trigonometric Functions in Motion
Trigonometric functions, like sine and cosine, play a vital role in describing harmonic motion. These functions are periodic, making them ideal for modeling repetitive motions like that of a spring. The cosine function in the given displacement equation, \(x = 10 \cos t\), describes how the displacement varies over time.

Cosine functions periodically oscillate between -1 and 1, capturing the full range from the spring's extreme positions. The amplitude of 10 indicates the maximum displacement the spring achieves.
  • The displacement starts at maximum when the cosine function is \(1\).
  • As time progresses, the displacement decreases as \(\cos t\) moves towards \(0\).
  • Eventually, it switches into negative values as \(\cos t\) becomes negative, indicating the spring has passed the equilibrium position into the opposite direction.
Similarly, the sine function, involved in the velocity calculation \(v(t) = -10 \sin t\), shows us when the spring reaches its peak speed and when it's momentarily at rest. Sine's nature of starting at zero complements cosine, beginning where cosine has its maximum or minimum.

For those studying oscillations, understanding how trigonometric functions govern movement is key. They explain everything from the spring's position to the intricacies of its speed, forming a foundation for analyzing many types of periodic motions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An explosion at an oil rig located in gulf waters causes an elliptical oil slick to spread on the surface from the rig. The slick is a constant \(20 \mathrm{cm}\) thick. After several days, when the major axis of the slick is \(2 \mathrm{km}\) long and the minor axis is \(3 / 4 \mathrm{km}\) wide, it is determined that its length is increasing at the rate of \(9 \mathrm{m} / \mathrm{h}\), and its width is increasing at the rate of \(3 \mathrm{m} / \mathrm{h}\). At what rate (in cubic meters per hour) is oil flowing from the site of the rig at that time?

The amount of work done by the heart's main pumping chamber, the left ventricle, is given by the equation $$W=P V+\frac{V \delta v^{2}}{2 g}$$, where \(W\) is the work per unit time, \(P\) is the average blood pressure, \(V\) is the volume of blood pumped out during the unit of time, \(\delta(\text { "delta") is the weight density of the blood, } v\) is the average velocity of the exiting blood, and \(g\) is the acceleration of gravity. When \(P, V, \delta,\) and \(v\) remain constant, \(W\) becomes a function of \(g,\) and the equation takes the simplified form $$ W=a+\frac{b}{g}(a, b \text { constant }) $$ As a member of NASA's medical team, you want to know how sensitive \(W\) is to apparent changes in \(g\) caused by flight maneuvers, and this depends on the initial value of \(g\). As part of your investigation, you decide to compare the effect on \(W\) of a given change \(d g\) on the moon, where \(g=1.6 \mathrm{m} / \mathrm{s}^{2}\), with the effect the same change \(d g\) would have on Earth, where \(g=9.8 \mathrm{m} / \mathrm{s}^{2} .\) Use the simplified equation above to find the ratio of \(d W_{\text {moon }}\) to \(d W_{\text {Earth }}\)

Use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval \(I\). Perform the following steps: a. Plot the function \(f\) over \(I\) b. Find the linearization \(L\) of the function at the point \(a\) c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta>0\) as you can, satisfying \(|x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon\) for \(\epsilon=0.5,0.1,\) and \(0.01 .\) Then check graphically to see if your \(\delta\) -estimate holds true. $$f(x)=\frac{x-1}{4 x^{2}+1}, \quad\left[-\frac{3}{4}, 1\right], \quad a=\frac{1}{2}$$

Two ships are steaming straight away from a point \(O\) along routes that make a \(120^{\circ}\) angle. Ship \(A\) moves at 14 knots (nautical miles per hour; a nautical mile is \(1852 \mathrm{m}\) ). Ship \(B\) moves at 21 knots. How fast are the ships moving apart when \(O A=5\) and \(O B=3\) nautical miles?

Give the position function \(s=f(t)\) of an object moving along the \(s\) -axis as a function of time \(t\). Graph \(f\) together with the velocity function \(v(t)=d s / d t=f^{\prime}(t)\) and the acceleration function \(a(t)=d^{2} s / d t^{2}=f^{\prime \prime}(t) .\) Comment on the object's behavior in relation to the signs and values of \(v\) and \(a .\) Include in your commentary such topics as the following: a. When is the object momentarily at rest? b. When does it move to the left (down) or to the right (up)? c. When does it change direction? d. When does it speed up and slow down? e. When is it moving fastest (highest speed)? Slowest? f. When is it farthest from the axis origin? $$s=t^{3}-6 t^{2}+7 t, \quad 0 \leq t \leq 4$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.