A Bobbing Cork A cork floating in a lake is bobbing in simple harmonic motion. Its displacement above the bottom of the lake is modeled by
where y is measured in meters and t is measured in minutes.
- (a) Find the frequency of the motion of the cork.
- (b) Sketch a graph of y.
- (c) Find the maximum displacement of the cork above the lake bottom.
(a)
The frequency for the motion of the cork floating in a lake.
Answer to Problem 27E
The frequency of a cork in simple harmonic motion is 10 cycles per minute.
Explanation of Solution
Given:
The equation for the displacement above the bottom of the lake is,
Where, y is in meters and t is in minutes.
Definition:
The equation for the simple harmonic motion which describes the displacement y of an object at time t is,
Where, A is the amplitude, b is the phase and k is the constant.
Calculation:
The equation for the displacement of the cork is,
The general equation for the displacement is,
Compare both the equation (1) and (2), the value of k is
The formula to calculate frequency of the cork is,
Substitute the value
Thus, the frequency of the cork is 10 cycles per minute.
(b)
To sketch: The graph for the cosine curve
Explanation of Solution
Sketch the graph for the cosine curve
Figure (1)
From the Figure (1), it is showed that graph is the cosine curve with oscillation between positive amplitude
(c)
The maximum displacement of the cork above the bottom of the lake.
Answer to Problem 27E
The maximum displacement for the cork is
Explanation of Solution
The amplitude of the cork is 0.2.
The equation for the displacement of the cork is,
For the maximum displacement of the cork, the cosine function must be maximum,
So, the cosine function is maximum at
Substitute 0 for t in equation (3),
Solve the equation,
Thus, the value of maximum displacement is
Chapter 5 Solutions
Precalculus: Mathematics for Calculus - 6th Edition
- 4c Consider the function f(x) = 10x + 4x5 - 4x³- 1. Enter the general antiderivative of f(x)arrow_forwardA tank contains 60 kg of salt and 2000 L of water. Pure water enters a tank at the rate 8 L/min. The solution is mixed and drains from the tank at the rate 11 L/min. Let y be the number of kg of salt in the tank after t minutes. The differential equation for this situation would be: dy dt y(0) =arrow_forwardSolve the initial value problem: y= 0.05y + 5 y(0) = 100 y(t) =arrow_forward
- y=f'(x) 1 8 The function f is defined on the closed interval [0,8]. The graph of its derivative f' is shown above. How many relative minima are there for f(x)? O 2 6 4 00arrow_forward60! 5!.7!.15!.33!arrow_forward• • Let > be a potential for the vector field F = (−2 y³, −6 xy² − 4 z³, −12 yz² + 4 2). Then the value of sin((-1.63, 2.06, 0.57) – (0,0,0)) is - 0.336 -0.931 -0.587 0.440 0.902 0.607 -0.609 0.146arrow_forward
- The value of cos(4M) where M is the magnitude of the vector field with potential ƒ = e² sin(лy) cos(π²) at x = 1, y = 1/4, z = 1/3 is 0.602 -0.323 0.712 -0.816 0.781 0.102 0.075 0.013arrow_forwardThere is exactly number a and one number b such that the vector field F = conservative. For those values of a and b, the value of cos(a) + sin(b) is (3ay + z, 3ayz + 3x, −by² + x) is -0.961 -0.772 -1.645 0.057 -0.961 1.764 -0.457 0.201arrow_forwardA: Tan Latitude / Tan P A = Tan 04° 30'/ Tan 77° 50.3' A= 0.016960 803 S CA named opposite to latitude, except when hour angle between 090° and 270°) B: Tan Declination | Sin P B Tan 052° 42.1'/ Sin 77° 50.3' B = 1.34 2905601 SCB is alway named same as declination) C = A + B = 1.35 9866404 S CC correction, A+/- B: if A and B have same name - add, If different name- subtract) = Tan Azimuth 1/Ccx cos Latitude) Tan Azimuth = 0.737640253 Azimuth = S 36.4° E CAzimuth takes combined name of C correction and Hour Angle - If LHA is between 0° and 180°, it is named "west", if LHA is between 180° and 360° it is named "east" True Azimuth= 143.6° Compass Azimuth = 145.0° Compass Error = 1.4° West Variation 4.0 East Deviation: 5.4 Westarrow_forward
- ds 5. Find a solution to this initial value problem: 3t2, s(0) = 5. dt 6. Find a solution to this initial value problem: A' = 0.03A, A(0) = 100.arrow_forward2) Drive the frequency responses of the following rotor system with Non-Symmetric Stator. The system contains both external and internal damping. Show that the system loses the reciprocity property.arrow_forward1) Show that the force response of a MDOF system with general damping can be written as: X liax) -Σ = ral iw-s, + {0} iw-s,arrow_forward
- Calculus: Early TranscendentalsCalculusISBN:9781285741550Author:James StewartPublisher:Cengage LearningThomas' Calculus (14th Edition)CalculusISBN:9780134438986Author:Joel R. Hass, Christopher E. Heil, Maurice D. WeirPublisher:PEARSONCalculus: Early Transcendentals (3rd Edition)CalculusISBN:9780134763644Author:William L. Briggs, Lyle Cochran, Bernard Gillett, Eric SchulzPublisher:PEARSON
- Calculus: Early TranscendentalsCalculusISBN:9781319050740Author:Jon Rogawski, Colin Adams, Robert FranzosaPublisher:W. H. FreemanCalculus: Early Transcendental FunctionsCalculusISBN:9781337552516Author:Ron Larson, Bruce H. EdwardsPublisher:Cengage Learning