Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals4h 44m
- 9. Graphical Applications of Integrals2h 27m
- 10. Physics Applications of Integrals 2h 22m
3. Techniques of Differentiation
The Chain Rule
Problem 3.7.98c
Textbook Question
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 when the mass hangs at rest. Suppose you push the mass to a position 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 , where is a constant measuring the stiffness of the spring (the larger the value of , the stiffer the spring) and 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 ( is increased by a factor of )?

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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