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Ch 15: Oscillations
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoKnight Calc 5th EditionCh 15: OscillationsProblema 53
Capitolo 15, Problema 53

A mass hanging from a spring oscillates with a period of 0.35 s. Suppose the mass and spring are swung in a horizontal circle, with the free end of the spring at the pivot. What rotation frequency, in rpm, will cause the spring's length to stretch by 15%?

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Determine the spring constant (k) using the oscillation period formula for a mass-spring system: \( T = 2\pi \sqrt{\frac{m}{k}} \). Rearrange to solve for \( k \): \( k = \frac{4\pi^2 m}{T^2} \). Substitute the given period \( T = 0.35 \ \text{s} \). Note that the mass \( m \) is not provided explicitly, so keep it as a variable for now.
When the spring stretches by 15%, the new length of the spring is \( L_{\text{new}} = 1.15L_0 \), where \( L_0 \) is the original length. The centripetal force required for circular motion is provided by the spring's restoring force: \( F_{\text{centripetal}} = F_{\text{spring}} \). Express this as \( m\omega^2L_{\text{new}} = k(L_{\text{new}} - L_0) \).
Substitute \( L_{\text{new}} = 1.15L_0 \) into the equation \( m\omega^2L_{\text{new}} = k(L_{\text{new}} - L_0) \). This simplifies to \( m\omega^2(1.15L_0) = k(0.15L_0) \). Cancel \( L_0 \) from both sides, resulting in \( m\omega^2(1.15) = k(0.15) \). Rearrange to solve for \( \omega \): \( \omega = \sqrt{\frac{k(0.15)}{m(1.15)}} \).
Substitute \( k = \frac{4\pi^2 m}{T^2} \) into the expression for \( \omega \): \( \omega = \sqrt{\frac{(4\pi^2 m / T^2)(0.15)}{m(1.15)}} \). Simplify by canceling \( m \): \( \omega = \sqrt{\frac{4\pi^2(0.15)}{T^2(1.15)}} \).
Convert \( \omega \) (angular velocity in radians per second) to rotation frequency in revolutions per minute (rpm). Use the relationship \( f = \frac{\omega}{2\pi} \) to find the frequency in Hz, then multiply by 60 to convert to rpm: \( \text{Frequency (rpm)} = \frac{\omega}{2\pi} \times 60 \). Substitute the expression for \( \omega \) into this formula to find the final frequency in rpm.

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Simple Harmonic Motion (SHM)

Simple Harmonic Motion describes the oscillatory motion of a mass attached to a spring, where the restoring force is proportional to the displacement from the equilibrium position. The period of oscillation is determined by the mass and the spring constant, and it is independent of the amplitude of the motion. Understanding SHM is crucial for analyzing how the mass behaves when subjected to forces, such as when it is swung in a circle.
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Simple Harmonic Motion of Pendulums

Centripetal Force

Centripetal Force is the net force required to keep an object moving in a circular path, directed towards the center of the circle. In this scenario, as the mass is swung in a horizontal circle, the tension in the spring provides the necessary centripetal force to maintain circular motion. The relationship between the mass, velocity, and radius of the circular path is essential for determining the conditions under which the spring stretches.
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Intro to Centripetal Forces

Spring Constant and Hooke's Law

Hooke's Law states that the force exerted by a spring is directly proportional to its extension or compression, expressed as F = kx, where k is the spring constant and x is the displacement from the equilibrium position. When the mass is swung in a circle, the spring stretches due to the centripetal force, and understanding the spring constant is vital for calculating how much the spring will stretch under the applied forces. This concept is key to determining the new length of the spring when it is subjected to rotational motion.
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Spring Force (Hooke's Law)
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