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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 3 - DerivativesProblem 3.7.98c
Chapter 3, Problem 3.7.98c

Vibrations of a spring Suppose an object of mass m is attached to the end of a spring hanging from the ceiling. The mass is at its equilibrium position y=0y=0 when the mass hangs at rest. Suppose you push the mass to a position y0y_0 units above its equilibrium position and release it. As the mass oscillates up and down (neglecting any friction in the system), the position y of the mass after t seconds is y=y0cos(tkm)y=y_0\(\cos\]\left\)(t\(\sqrt{\frac{k}{m}\)}\(\right\)), where k>0k>0 is a constant measuring the stiffness of the spring (the larger the value of kk, the stiffer the spring) and yy is positive in the upward direction.

Use equation (4) to answer the following questions.
c. How would the velocity be affected if the experiment were repeated with a spring having four times the stiffness (kk is increased by a factor of 44)?

Verified step by step guidance
1
Step 1: Start by recalling the position function of the mass on the spring, which is given by y = y_0 \(\cos\)\(\left\)(t\(\sqrt{\frac{k}{m}\)}\(\right\)).
Step 2: To find the velocity, differentiate the position function with respect to time t. Use the chain rule for differentiation: \(\frac{dy}{dt}\) = -y_0 \(\sin\[\left\)(t\(\sqrt{\frac{k}{m}\)}\(\right\)) \(\cdot\) \(\frac{d}{dt}\]\left\)(t\(\sqrt{\frac{k}{m}\)}\(\right\)).
Step 3: Differentiate the inner function t\(\sqrt{\frac{k}{m}\)} with respect to t, which results in \(\sqrt{\frac{k}{m}\)}.
Step 4: Substitute the derivative of the inner function back into the velocity expression: \(\frac{dy}{dt}\) = -y_0 \(\sin\)\(\left\)(t\(\sqrt{\frac{k}{m}\)}\(\right\)) \(\cdot\) \(\sqrt{\frac{k}{m}\)}.
Step 5: Consider the effect of increasing the stiffness k by a factor of 4. Substitute 4k for k in the velocity expression: \(\frac{dy}{dt}\) = -y_0 \(\sin\)\(\left\)(t\(\sqrt{\frac{4k}{m}\)}\(\right\)) \(\cdot\) \(\sqrt{\frac{4k}{m}\)}. Simplify the expression to see how the velocity changes.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Simple Harmonic Motion (SHM)

Simple Harmonic Motion is a type of periodic motion where an object oscillates around an equilibrium position. In the context of a mass-spring system, the motion is characterized by a restoring force proportional to the displacement from equilibrium, leading to sinusoidal motion. The position of the mass can be described using trigonometric functions, such as cosine, which reflects the oscillatory nature of the system.
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Spring Constant (k)

The spring constant, denoted as 'k', is a measure of a spring's stiffness. It quantifies the force required to stretch or compress the spring by a unit distance. A higher value of 'k' indicates a stiffer spring, which results in faster oscillations and greater restoring force when the spring is displaced from its equilibrium position.
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Velocity in SHM

In Simple Harmonic Motion, the velocity of the oscillating object varies with time and is derived from the position function. The velocity can be expressed as the derivative of the position with respect to time. When the spring constant 'k' increases, the angular frequency of the motion increases, leading to a higher maximum velocity during oscillation, as the object moves more quickly through its equilibrium position.
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Related Practice
Textbook Question

{Use of Tech} Power and energy Power and energy are often used interchangeably, but they are quite different. Energy is what makes matter move or heat up. It is measured in units of joules or Calories, where 1 Cal=4184 J. One hour of walking consumes roughly 10⁶J, or 240 Cal. On the other hand, power is the rate at which energy is used, which is measured in watts, where 1 W=1 J/s. Other useful units of power are kilowatts (1 kW=10³ W) and megawatts (1 MW=10⁶ W). If energy is used at a rate of 1 kW for one hour, the total amount of energy used is 1 kilowatt-hour (1 kWh=3.6×10⁶ J) Suppose the cumulative energy used in a large building over a 24-hr period is given by E(t)=100t+4t²−t³ / 9kWh where t=0 corresponds to midnight.

c. Graph the power function and interpret the graph. What are the units of power in this case?

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Textbook Question

Derivatives of inverse functions from a table Use the following tables to determine the indicated derivatives or state that the derivative cannot be determined. <IMAGE>

c. (f^-1)'(1)

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Textbook Question

Another way to approximate derivatives is to use the centered difference quotient: f' (a) ≈ f(a+h) - f(a- h) / 2h. Again, consider f(x) = √x.

c. Explain why it is not necessary to use negative values of h in the table of part (b).

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Textbook Question

62–65. {Use of Tech} Graphing f and f'

c. Verify that the zeros of f' correspond to points at which f has a horizontal tangent line.

f(x)=(sec^−1 x)/x on [1,∞)

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Textbook Question

Derivatives of sin^n x Calculate the following derivatives using the Product Rule.

c. d/dx (sin⁴ x)

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Suppose a stone is thrown vertically upward from the edge of a cliff on Earth with an initial velocity of 19.6 m/s from a height of 24.5 m above the ground. The height (in meters) of the stone above the ground t seconds after it is thrown is s(t) = -4.9t²+19.6t+24.5.

c. What is the height of the stone at the highest point?

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