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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
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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
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