Ch 15: Oscillations

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 15: Oscillations

Problem 53

Chapter 15, Problem 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%?

Verified step by step guidance

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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Key Concepts

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

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.

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.

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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Related Practice

Textbook Question

A 200 g block hangs from a spring with spring constant 10 N/m. At t = 0 s the block is 20 cm below the equilibrium point and moving upward with a speed of 100 cm/s. What are the block's distance from equilibrium when the speed is 50 cm/s?

Textbook Question

A 500 g wood block on a frictionless table is attached to a horizontal spring. A 50 g dart is shot into the face of the block opposite the spring, where it sticks. Afterward, the spring oscillates with a period of 1.5 s and an amplitude of 20 cm. How fast was the dart moving when it hit the block?

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Textbook Question

A 1.00 kg block is attached to a horizontal spring with spring constant 2500 N/m. The block is at rest on a frictionless surface. A 10 g bullet is fired into the block, in the face opposite the spring, and sticks. What was the bullet's speed if the subsequent oscillations have an amplitude of 10.0 cm?

Textbook Question

Interestingly, there have been several studies using cadavers to determine the moments of inertia of human body parts, information that is important in biomechanics. In one study, the center of mass of a 5.0 kg lower leg was found to be 18 cm from the knee. When the leg was allowed to pivot at the knee and swing freely as a pendulum, the oscillation frequency was 1.6 Hz. What was the moment of inertia of the lower leg about the knee joint?

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Textbook Question

A compact car has a mass of 1200 kg. Assume that the car has one spring on each wheel, that the springs are identical, and that the mass is equally distributed over the four springs. What will be the car's oscillation frequency while carrying four 70 kg passengers?

Textbook Question

Scientists are measuring the properties of a newly discovered elastic material. They create a 1.5-m-long, 1.6-mm-diameter cord, attach an 850 g mass to the lower end, then pull the mass down 2.5 mm and release it. Their high-speed video camera records 36 oscillations in 2.0 s. What is Young's modulus of the material?