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Ch 14: Periodic Motion
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 14th EditionCh 14: Periodic MotionProblem 30
Chapter 14, Problem 30

A cheerleader waves her pom-pom in SHM with an amplitude of 18.0 cm and a frequency of 0.850 Hz. Find (a) the maximum magnitude of the acceleration and of the velocity; (b) the acceleration and speed when the pom-pom's coordinate is x = +9.0 cm; (c) the time required to move from the equilibrium position directly to a point 12.0 cm away. (d) Which of the quantities asked for in parts (a), (b), and (c) can be found by using the energy approach used in Section 14.3, and which cannot? Explain.

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To solve part (a), we need to find the maximum velocity and acceleration in simple harmonic motion (SHM). The maximum velocity (v_max) is given by the formula v_max = ωA, where ω is the angular frequency and A is the amplitude. First, calculate the angular frequency using ω = 2πf, where f is the frequency. Then, use the amplitude A = 18.0 cm to find v_max.
For the maximum acceleration (a_max), use the formula a_max = ω²A. With the angular frequency ω already calculated, substitute it and the amplitude A = 18.0 cm into the formula to find a_max.
In part (b), to find the acceleration and speed when the pom-pom's coordinate is x = +9.0 cm, use the equations for velocity and acceleration in SHM. The velocity v at position x is given by v = ±ω√(A² - x²). Calculate this using the known values of ω, A, and x. The acceleration a is given by a = -ω²x. Substitute the values of ω and x to find the acceleration.
For part (c), to find the time required to move from the equilibrium position to a point 12.0 cm away, use the equation for displacement in SHM: x(t) = A cos(ωt + φ). At the equilibrium position, φ = 0, so x(t) = A cos(ωt). Solve for t when x = 12.0 cm.
In part (d), consider which quantities can be found using the energy approach. The energy approach involves using the conservation of mechanical energy in SHM, where the total energy E = (1/2)mv² + (1/2)kx² is constant. Maximum velocity and acceleration can be found using energy considerations, as they relate to kinetic and potential energy. However, the time to move to a specific position is not directly found using energy methods, as it involves kinematic equations.

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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 the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement. It is characterized by sinusoidal oscillations, with key parameters including amplitude, frequency, and phase. Understanding SHM is crucial for analyzing the motion of the pom-pom, as it dictates the behavior of velocity and acceleration over time.
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Kinematics of SHM

In SHM, the velocity and acceleration of an object can be described using trigonometric functions. The maximum velocity occurs at the equilibrium position, while maximum acceleration occurs at the maximum displacement. The velocity v(t) is given by v_max * cos(ωt + φ), and acceleration a(t) is given by -ω² * x(t), where ω is the angular frequency. These equations are essential for calculating the maximum values and specific values at given displacements.
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Energy in SHM

The energy approach in SHM involves the conservation of mechanical energy, where the total energy is the sum of kinetic and potential energy. At maximum displacement, potential energy is maximized, while kinetic energy is zero, and vice versa at the equilibrium position. This approach can be used to find quantities like maximum speed and acceleration, but not the time to reach a specific displacement, which requires kinematic equations.
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Related Practice
Textbook Question

You pull a simple pendulum 0.240 m long to the side through an angle of 3.50° and release it. How much time does it take the pendulum bob to reach its highest speed?

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Textbook Question

A 0.500-kg glider, attached to the end of an ideal spring with force constant k = 450 N/m, undergoes SHM with an amplitude of 0.040 m. Compute the speed of the glider when it is at x = -0.015 m.

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Textbook Question

A 0.500-kg glider, attached to the end of an ideal spring with force constant k = 450 N/m, undergoes SHM with an amplitude of 0.040 m. Compute the total mechanical energy of the glider at any point in its motion

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Textbook Question

A thrill-seeking cat with mass 4.00 kg is attached by a harness to an ideal spring of negligible mass and oscillates vertically in SHM. The amplitude is 0.050 m, and at the highest point of the motion the spring has its natural unstretched length. Calculate the elastic potential energy of the spring (take it to be zero for the unstretched spring), the kinetic energy of the cat, the gravitational potential energy of the system relative to the lowest point of the motion, and the sum of these three energies when the cat is at its highest point.

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Textbook Question

A mass is oscillating with amplitude A at the end of a spring. How far (in terms of A) is this mass from the equilibrium position of the spring when the elastic potential energy equals the kinetic energy?

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Textbook Question

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. The amplitude of the motion is 0.250 m and the period is 3.20 s. What are the speed and acceleration of the block when x = 0.160 m?

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