College Physics:
College Physics:
11th Edition
ISBN: 9781305965515
Author: SERWAY, Raymond A.
Publisher: Brooks/Cole Pub Co
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Chapter 13, Problem 32P

A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 7.50 N is applied. A 0.500-kg particle rests on a frictionless horizontal surface and is attached to the free end of the spring. The particle is displaced from the origin to x = 5.00 cm and released from rest at t = 0. (a) What is the force constant of the spring? (b) What are the angular frequency ω, the frequency, and the period of the motion? (c) What is the total energy of the system? (d) What is the amplitude of the motion? (c) What are the maximum velocity and the maximum acceleration of the particle? (f) Determine the displacement x of the particle from the equilibrium position at t = 0.500 s. (g) Determine the velocity and acceleration of the particle when t = 0.500 s.

(a)

Expert Solution
Check Mark
To determine
The force constant of the spring.

Answer to Problem 32P

The force constant of the spring is 250Nm-1 .

Explanation of Solution

Given info: The force applied F=7.50N . The spring is stretched to 3.00cm .

Explanation:

From the hook’s law, the restoring force is,

F=kx

  • F is the restoring force
  • k is the force constant of the spring
  • x is the displacement from the equilibrium position

The magnitude of the force constant will be,

k=Fx

Substituting F=7.50N and x=3.00cm

k=(7.50N)(3.00cm)=(7.50N)(3.00×102m)=250Nm-1

Conclusion: The force constant of the spring is 250Nm-1 .

(b)

Expert Solution
Check Mark
To determine
The angular frequency, the frequency, and the time period of the oscillation.

Answer to Problem 32P

The angular frequency is 22.4rads-1 , the frequency is 3.56Hz and the period of the oscillation is 0.281s .

Explanation of Solution

Given info: The mass of the particle is 0.500kg The particle is released at x=5.00cm from rest, so the amplitude of the oscillation is 5.00cm .

Explanation:

The angular frequency is defined as,

ω=km (1)

  • ω is the angular frequency
  • k is the force constant
  • m is the mass of the particle

Substituting m=0.500kg and from section (a) k=250Nm-1 ,

ω=(250Nm-1)(0.500kg)=22.4rads-1

The frequency of the oscillation is given by,

f=12πkm (2)

  • f is the frequency of the oscillation
  • k is the force constant
  • m is the mass of the particle

Substituting m=0.500kg and from section (a) k=250Nm-1 ,

f=12(250Nm-1)(0.500kg)=3.56Hz

The time period is defined by,

T=1f

Substituting f=3.56Hz ,

T=1(3.56Hz)=0.281s

Conclusion: The angular frequency is 22.4rads-1 , the frequency is 3.56Hz and the period of the oscillation is 0.281s .

(c)

Expert Solution
Check Mark
To determine
The total energy of the system.

Answer to Problem 32P

The total energy of the system is 0.313J

Explanation of Solution

Given info: The particle is displaced from the origin to x= 5.00cm and released from rest at t=0 .

Explanation:

The total energy of a simple harmonic system is

E=12kx2+12mv2

  • E is the total energy
  • k is the force constant
  • x is the displacement from equilibrium position
  • m is the mass of the particle
  • v is the velocity

Substituting x=5.00cm , v=0 and k=250Nm-1

E=12(250Nm-1)(5.00cm)2+0=0.313J

Conclusion: The total energy of the system is 0.313J .

(d)

Expert Solution
Check Mark
To determine
The amplitude of the system.

Answer to Problem 32P

The amplitude is 5.00cm .

Explanation of Solution

Given info: The force constant of the spring is 250Nm-1 . The total energy of the system is 0.313J .

Explanation:

The total energy of a simple harmonic system is stored as elastic potential energy when the system is in the turning points.

E=12kA2

  • E is the total energy
  • k is the force constant
  • A is the amplitude

On re-arranging for the amplitude

A=2Ek

Substituting k=250Nm-1 and E=0.313J

A=2(0.313J)(250Nm-1)=5.00×102m=5.00cm

Conclusion: The amplitude of the oscillation is 5.00cm .

(e)

Expert Solution
Check Mark
To determine
The maximum velocity and the maximum acceleration of the system.

Answer to Problem 32P

The maximum velocity of the system is 1.12ms-1 . The maximum acceleration of the system is 25.0ms-2 .

Explanation of Solution

Given info: The total energy of the system is 0.313J . The mass of the object is 0.500kg .

Explanation:

When the system is in equilibrium position, the total energy is in the form of kinetic energy,

E=12mvmax2

Re-arranging

vmax=2Em

Substituting m=0.500kg and E=0.313J

vmax=2(0.313J)(0.500kg)=1.12ms-1

The maximum force is given by,

Fmax=kA (1)

  • k is the force constant
  • A is the amplitude

Using Newton’s second law,

Fmax=mamax (2)

From (1) and (2)

amax=kAm

Substituting k=250Nm-1 A=5.00cm and m=0.500kg ,

amax=(250Nm-1)(5.00cm)(0.500kg)=25.0ms-2

Conclusion: The maximum velocity of the system is 1.12ms-1 . The maximum acceleration of the system is 25.0ms-2 .

(f)

Expert Solution
Check Mark
To determine
The displacement x of the particle from the equilibrium position t=0.500s .

Answer to Problem 32P

The displacement is 0.919cm .

Explanation of Solution

Given info: The mass of the object m=0.500kg . The force constant of the spring k=250Nm-1 . Amplitude A=5.00cm , time t=0.500s

Explanation:

The general expression of position as a function of time for an object moving in simple harmonic motion is given by,

x=Acos(ωt) (1)

  • x is the position
  • A is the amplitude of the simple harmonic motion
  • ω is the angular frequency
  • t is time taken

The angular frequency is defined as,

ω=km (2)

  • ω is the angular frequency
  • k is the force constant
  • m is the mass of the particle

From (1) and (2)

x=Acos((km)t)

Substituting m=0.500kg , k=250Nm-1 , A=5.00cm , t=0.500s

x=(5.00cm)cos(((250Nm-1)(0.500kg))(0.500s))=0.919cm

Conclusion: The displacement from the equilibrium position is 0.919cm .

(g)

Expert Solution
Check Mark
To determine
The velocity and acceleration of the particle at t=0.500s .

Answer to Problem 32P

The velocity is 1.10ms-1 and the acceleration is 4.59ms-2

Explanation of Solution

Given info: The mass of the object m=0.500kg . The force constant of the spring k=250Nm-1 . Amplitude A=5.00cm , time t=0.500s .

Explanation:

The general expression of position as a function of time for an object moving in simple harmonic motion is given by,

x=Acos(ωt) (1)

  • x is the position
  • A is the amplitude of the simple harmonic motion
  • ω is the angular frequency
  • t is time taken

To find the expression for velocity differentiate (1)

v=Aωsin(ωt) (2)

To find the expression for acceleration differentiate (1) twice

a=Aω2cos(ωt) (3)

The angular frequency is defined as,

ω=km (4)

  • ω is the angular frequency
  • k is the force constant
  • m is the mass of the particle

Hence,

v=A(km)sin((km)t)

a=A(km)2cos((km)t)

Substituting m=0.500kg , k=250Nm-1 , A=5.00cm , t=0.500s , the velocity will be,

v=(5.00cm)((250Nm-1)(0.500kg))sin((0.500s)((250Nm-1)(0.500kg)))=(5.00×102m)((250Nm-1)(0.500kg))sin((0.500s)((250Nm-1)(0.500kg)))=1.10ms-1

Similarly the acceleration,

a=(5.00cm)((250Nm-1)(0.500kg))2cos((0.500s)((250Nm-1)(0.500kg)))=(5.00×102m)((250Nm-1)(0.500kg))2sin((0.500s)((250Nm-1)(0.500kg)))=4.59ms-2

Conclusion: The velocity of the oscillator is 1.10ms-1 and the acceleration is 4.59ms-2 .

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Chapter 13 Solutions

College Physics:

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