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Please help solve for v in (m/s)

Please help solve for v in (m/s)

Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter12: Oscillatory Motion
Section: Chapter Questions
Problem 17P: A block of unknown mass is attached to a spring with a spring constant of 6.50 N/m and undergoes...
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Please help solve for v in (m/s)

A particle with mass 2.59 kg oscillates horizontally at the end of a horizontal spring. A student measures an amplitude of 0.897 m and a duration of 127 s for 71 cycles of oscillation. Find the frequency, f, the speed at the equilibrium position, vmax, the spring constant, k, the potential energy at an endpoint, Umax, the potential energy when the particle is located 45.1% of the amplitude away from the equiliibrium position, U, and the kinetic energy, K, and the speed, v, at the same position. f = 0.559 Hz Umax m/s k = 31.958 N/m U max 12.856 J U = J K = J V = m/s
Transcribed Image Text:A particle with mass 2.59 kg oscillates horizontally at the end of a horizontal spring. A student measures an amplitude of 0.897 m and a duration of 127 s for 71 cycles of oscillation. Find the frequency, f, the speed at the equilibrium position, vmax, the spring constant, k, the potential energy at an endpoint, Umax, the potential energy when the particle is located 45.1% of the amplitude away from the equiliibrium position, U, and the kinetic energy, K, and the speed, v, at the same position. f = 0.559 Hz Umax m/s k = 31.958 N/m U max 12.856 J U = J K = J V = m/s
Expert Solution
Step 1

Introduction:

The total energy of a spring block oscillator executing simple harmonic motion is

E =12kx2 + 12mv2

Maximum speed of the block is

vm= ωA

Speed of the block at a distance x from the equilibrium position is

v=ωA2-x2

Angular frequency is ω = km

Period of oscillation is T =2πω

Step 2

(given)

Mass of the particle (m) = 2.59 kg

Amplitude of the oscillation (A) = 0.897 m

n = 71 and t= 127 s

Frequency of oscillation = 0.559 Hz

Spring constant of the spring = 31.958 N/m

Maximum Potential energy  of the oscillator = U max  = 12.856 J

 

Angular frequency is

ω=2πf=km   = 2(3.14)(0.559)   =3.51 rad/s 

Instantaneous position of the particle is

x = 0.451× A

   = 0.451×0.897

   = 0.4045 m

 

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