To prove the relation for system.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 16, Problem 82PQ

To determine

To prove the relation for system.

Expert Solution & Answer

Answer to Problem 82PQ

The solution to the equation is $Fmaxcos(ωdrvt)=m(d2y/dt2)+b(dy/dt)+ky_$ .

Explanation of Solution

Write the expression for the given solution of the equation.

$y(t)=Adrvcos(ωdrvt+ϕdrv)$ (I)

Here, $y(t)$ is the instantaneous displacement, $Adrv$ is the amplitude, $ωdrv$ is the angular frequency, $t$ is the time and $ϕdrv$ is the phase angle.

Write the expression for the equation.

$Fmaxcos(ωdrvt)=m(d2y/dt2)+b(dy/dt)+ky$ (II)

Here, $Fmax$ is the maximum force, $(d2y/dt2)$ is the second derivative and $(dy/dt)$ is the first derivative.

Write the expression for the angular frequency.

$ωdrv=km$ (III)

Here, $k$ is wave vector and $m$ is mass.

Conclusion:

Taking first and second derivative from equation (I).

$dydt=−ωdrvAdrvsin(ωdrvt+ϕdrv)$ (IV)

$d2ydt2=−ωdrv2Adrvcos(ωdrvt+ϕdrv)$ (V)

Substitute $−ωdrvAdrvsin(ωdrvt+ϕdrv)$ for $dydt$ , $−ωdrv2Adrvcos(ωdrvt+ϕdrv)$ for $d2ydt2$ and $Adrvcos(ωdrvt+ϕdrv)$ for $y(t)$ in Equation (II) to find $Fmaxcos(ωdrvt)$ .

$Fmaxcos(ωdrvt)=(m[−ωdrv2Adrvcos(ωdrvt+ϕdrv)]+b[−ωdrvAdrvsin(ωdrvt+ϕdrv)]+k[Adrvcos(ωdrvt+ϕdrv)])$ (VI)

Using trigonometric identity and comparing both side the components of sine and cosine terms.

$Fmax=Adrv[(k−mωdrv2)cos(ϕdrv)−bωdrvsinϕdrv]$ (VII)

$0=Adrv[−(k−mωdrv2)sin(ϕdrv)−bωdrvcosϕdrv]$ (VIII)

Write the expression for the trigonometric identities by using equation (VIII).

$tan(ϕdrv)=−bωdrvk−mωdrv2$ (IX)

$sin(ϕdrv)=−bωdrv(bωdrv)2+(k−mωdrv2)2$ (X)

$cos(ϕdrv)=(k−mωdrv2)(bωdrv)2+(k−mωdrv2)2$ (XI)

Substitute $−bωdrv(bωdrv)2+(k−mωdrv2)2$ for $sin(ϕdrv)$ and $(k−mωdrv2)(bωdrv)2+(k−mωdrv2)2$ for $cos(ϕdrv)$ in equation (VII) to find $Adrv$ .

$Fmax=Adrv[(k−mωdrv2)[(k−mωdrv2)(bωdrv)2+(k−mωdrv2)2]−bωdrv[−bωdrv(bωdrv)2+(k−mωdrv2)2]]=Adrv(bωdrv)2+m2(ω2−ωdrv2)Adrv=Fmax(bωdrv)2+m2(ω2−ωdrv2)$

Thus, the solution is satisfying both side of the equation.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Make sure to draw a Free Body Diagram please as well

Make sure to draw a Free Body Diagram as well

Answer 2

Question

Chapter 16, Problem 82PQ

To determine

To prove the relation for system.

Expert Solution & Answer

Answer to Problem 82PQ

The solution to the equation is $Fmaxcos(ωdrvt)=m(d2y/dt2)+b(dy/dt)+ky_$ .

Explanation of Solution

Write the expression for the given solution of the equation.

$y(t)=Adrvcos(ωdrvt+ϕdrv)$ (I)

Here, $y(t)$ is the instantaneous displacement, $Adrv$ is the amplitude, $ωdrv$ is the angular frequency, $t$ is the time and $ϕdrv$ is the phase angle.

Write the expression for the equation.

$Fmaxcos(ωdrvt)=m(d2y/dt2)+b(dy/dt)+ky$ (II)

Here, $Fmax$ is the maximum force, $(d2y/dt2)$ is the second derivative and $(dy/dt)$ is the first derivative.

Write the expression for the angular frequency.

$ωdrv=km$ (III)

Here, $k$ is wave vector and $m$ is mass.

Conclusion:

Taking first and second derivative from equation (I).

$dydt=−ωdrvAdrvsin(ωdrvt+ϕdrv)$ (IV)

$d2ydt2=−ωdrv2Adrvcos(ωdrvt+ϕdrv)$ (V)

Substitute $−ωdrvAdrvsin(ωdrvt+ϕdrv)$ for $dydt$ , $−ωdrv2Adrvcos(ωdrvt+ϕdrv)$ for $d2ydt2$ and $Adrvcos(ωdrvt+ϕdrv)$ for $y(t)$ in Equation (II) to find $Fmaxcos(ωdrvt)$ .

$Fmaxcos(ωdrvt)=(m[−ωdrv2Adrvcos(ωdrvt+ϕdrv)]+b[−ωdrvAdrvsin(ωdrvt+ϕdrv)]+k[Adrvcos(ωdrvt+ϕdrv)])$ (VI)

Using trigonometric identity and comparing both side the components of sine and cosine terms.

$Fmax=Adrv[(k−mωdrv2)cos(ϕdrv)−bωdrvsinϕdrv]$ (VII)

$0=Adrv[−(k−mωdrv2)sin(ϕdrv)−bωdrvcosϕdrv]$ (VIII)

Write the expression for the trigonometric identities by using equation (VIII).

$tan(ϕdrv)=−bωdrvk−mωdrv2$ (IX)

$sin(ϕdrv)=−bωdrv(bωdrv)2+(k−mωdrv2)2$ (X)

$cos(ϕdrv)=(k−mωdrv2)(bωdrv)2+(k−mωdrv2)2$ (XI)

Substitute $−bωdrv(bωdrv)2+(k−mωdrv2)2$ for $sin(ϕdrv)$ and $(k−mωdrv2)(bωdrv)2+(k−mωdrv2)2$ for $cos(ϕdrv)$ in equation (VII) to find $Adrv$ .

$Fmax=Adrv[(k−mωdrv2)[(k−mωdrv2)(bωdrv)2+(k−mωdrv2)2]−bωdrv[−bωdrv(bωdrv)2+(k−mωdrv2)2]]=Adrv(bωdrv)2+m2(ω2−ωdrv2)Adrv=Fmax(bωdrv)2+m2(ω2−ωdrv2)$

Thus, the solution is satisfying both side of the equation.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 16, Problem 82PQ

To determine

To prove the relation for system.

Expert Solution & Answer

Answer to Problem 82PQ

The solution to the equation is $Fmaxcos(ωdrvt)=m(d2y/dt2)+b(dy/dt)+ky_$ .

Explanation of Solution

Write the expression for the given solution of the equation.

$y(t)=Adrvcos(ωdrvt+ϕdrv)$ (I)

Here, $y(t)$ is the instantaneous displacement, $Adrv$ is the amplitude, $ωdrv$ is the angular frequency, $t$ is the time and $ϕdrv$ is the phase angle.

Write the expression for the equation.

$Fmaxcos(ωdrvt)=m(d2y/dt2)+b(dy/dt)+ky$ (II)

Here, $Fmax$ is the maximum force, $(d2y/dt2)$ is the second derivative and $(dy/dt)$ is the first derivative.

Write the expression for the angular frequency.

$ωdrv=km$ (III)

Here, $k$ is wave vector and $m$ is mass.

Conclusion:

Taking first and second derivative from equation (I).

$dydt=−ωdrvAdrvsin(ωdrvt+ϕdrv)$ (IV)

$d2ydt2=−ωdrv2Adrvcos(ωdrvt+ϕdrv)$ (V)

Substitute $−ωdrvAdrvsin(ωdrvt+ϕdrv)$ for $dydt$ , $−ωdrv2Adrvcos(ωdrvt+ϕdrv)$ for $d2ydt2$ and $Adrvcos(ωdrvt+ϕdrv)$ for $y(t)$ in Equation (II) to find $Fmaxcos(ωdrvt)$ .

$Fmaxcos(ωdrvt)=(m[−ωdrv2Adrvcos(ωdrvt+ϕdrv)]+b[−ωdrvAdrvsin(ωdrvt+ϕdrv)]+k[Adrvcos(ωdrvt+ϕdrv)])$ (VI)

Using trigonometric identity and comparing both side the components of sine and cosine terms.

$Fmax=Adrv[(k−mωdrv2)cos(ϕdrv)−bωdrvsinϕdrv]$ (VII)

$0=Adrv[−(k−mωdrv2)sin(ϕdrv)−bωdrvcosϕdrv]$ (VIII)

Write the expression for the trigonometric identities by using equation (VIII).

$tan(ϕdrv)=−bωdrvk−mωdrv2$ (IX)

$sin(ϕdrv)=−bωdrv(bωdrv)2+(k−mωdrv2)2$ (X)

$cos(ϕdrv)=(k−mωdrv2)(bωdrv)2+(k−mωdrv2)2$ (XI)

Substitute $−bωdrv(bωdrv)2+(k−mωdrv2)2$ for $sin(ϕdrv)$ and $(k−mωdrv2)(bωdrv)2+(k−mωdrv2)2$ for $cos(ϕdrv)$ in equation (VII) to find $Adrv$ .

$Fmax=Adrv[(k−mωdrv2)[(k−mωdrv2)(bωdrv)2+(k−mωdrv2)2]−bωdrv[−bωdrv(bωdrv)2+(k−mωdrv2)2]]=Adrv(bωdrv)2+m2(ω2−ωdrv2)Adrv=Fmax(bωdrv)2+m2(ω2−ωdrv2)$

Thus, the solution is satisfying both side of the equation.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 16, Problem 82PQ

To determine

To prove the relation for system.

Expert Solution & Answer

Answer to Problem 82PQ

The solution to the equation is $Fmaxcos(ωdrvt)=m(d2y/dt2)+b(dy/dt)+ky_$ .

Explanation of Solution

Write the expression for the given solution of the equation.

$y(t)=Adrvcos(ωdrvt+ϕdrv)$ (I)

Here, $y(t)$ is the instantaneous displacement, $Adrv$ is the amplitude, $ωdrv$ is the angular frequency, $t$ is the time and $ϕdrv$ is the phase angle.

Write the expression for the equation.

$Fmaxcos(ωdrvt)=m(d2y/dt2)+b(dy/dt)+ky$ (II)

Here, $Fmax$ is the maximum force, $(d2y/dt2)$ is the second derivative and $(dy/dt)$ is the first derivative.

Write the expression for the angular frequency.

$ωdrv=km$ (III)

Here, $k$ is wave vector and $m$ is mass.

Conclusion:

Taking first and second derivative from equation (I).

$dydt=−ωdrvAdrvsin(ωdrvt+ϕdrv)$ (IV)

$d2ydt2=−ωdrv2Adrvcos(ωdrvt+ϕdrv)$ (V)

Substitute $−ωdrvAdrvsin(ωdrvt+ϕdrv)$ for $dydt$ , $−ωdrv2Adrvcos(ωdrvt+ϕdrv)$ for $d2ydt2$ and $Adrvcos(ωdrvt+ϕdrv)$ for $y(t)$ in Equation (II) to find $Fmaxcos(ωdrvt)$ .

$Fmaxcos(ωdrvt)=(m[−ωdrv2Adrvcos(ωdrvt+ϕdrv)]+b[−ωdrvAdrvsin(ωdrvt+ϕdrv)]+k[Adrvcos(ωdrvt+ϕdrv)])$ (VI)

Using trigonometric identity and comparing both side the components of sine and cosine terms.

$Fmax=Adrv[(k−mωdrv2)cos(ϕdrv)−bωdrvsinϕdrv]$ (VII)

$0=Adrv[−(k−mωdrv2)sin(ϕdrv)−bωdrvcosϕdrv]$ (VIII)

Write the expression for the trigonometric identities by using equation (VIII).

$tan(ϕdrv)=−bωdrvk−mωdrv2$ (IX)

$sin(ϕdrv)=−bωdrv(bωdrv)2+(k−mωdrv2)2$ (X)

$cos(ϕdrv)=(k−mωdrv2)(bωdrv)2+(k−mωdrv2)2$ (XI)

Substitute $−bωdrv(bωdrv)2+(k−mωdrv2)2$ for $sin(ϕdrv)$ and $(k−mωdrv2)(bωdrv)2+(k−mωdrv2)2$ for $cos(ϕdrv)$ in equation (VII) to find $Adrv$ .

$Fmax=Adrv[(k−mωdrv2)[(k−mωdrv2)(bωdrv)2+(k−mωdrv2)2]−bωdrv[−bωdrv(bωdrv)2+(k−mωdrv2)2]]=Adrv(bωdrv)2+m2(ω2−ωdrv2)Adrv=Fmax(bωdrv)2+m2(ω2−ωdrv2)$

Thus, the solution is satisfying both side of the equation.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 16, Problem 82PQ

To determine

To prove the relation for system.

Expert Solution & Answer

Answer to Problem 82PQ

The solution to the equation is $Fmaxcos(ωdrvt)=m(d2y/dt2)+b(dy/dt)+ky_$ .

Explanation of Solution

Write the expression for the given solution of the equation.

$y(t)=Adrvcos(ωdrvt+ϕdrv)$ (I)

Here, $y(t)$ is the instantaneous displacement, $Adrv$ is the amplitude, $ωdrv$ is the angular frequency, $t$ is the time and $ϕdrv$ is the phase angle.

Write the expression for the equation.

$Fmaxcos(ωdrvt)=m(d2y/dt2)+b(dy/dt)+ky$ (II)

Here, $Fmax$ is the maximum force, $(d2y/dt2)$ is the second derivative and $(dy/dt)$ is the first derivative.

Write the expression for the angular frequency.

$ωdrv=km$ (III)

Here, $k$ is wave vector and $m$ is mass.

Conclusion:

Taking first and second derivative from equation (I).

$dydt=−ωdrvAdrvsin(ωdrvt+ϕdrv)$ (IV)

$d2ydt2=−ωdrv2Adrvcos(ωdrvt+ϕdrv)$ (V)

Substitute $−ωdrvAdrvsin(ωdrvt+ϕdrv)$ for $dydt$ , $−ωdrv2Adrvcos(ωdrvt+ϕdrv)$ for $d2ydt2$ and $Adrvcos(ωdrvt+ϕdrv)$ for $y(t)$ in Equation (II) to find $Fmaxcos(ωdrvt)$ .

$Fmaxcos(ωdrvt)=(m[−ωdrv2Adrvcos(ωdrvt+ϕdrv)]+b[−ωdrvAdrvsin(ωdrvt+ϕdrv)]+k[Adrvcos(ωdrvt+ϕdrv)])$ (VI)

Using trigonometric identity and comparing both side the components of sine and cosine terms.

$Fmax=Adrv[(k−mωdrv2)cos(ϕdrv)−bωdrvsinϕdrv]$ (VII)

$0=Adrv[−(k−mωdrv2)sin(ϕdrv)−bωdrvcosϕdrv]$ (VIII)

Write the expression for the trigonometric identities by using equation (VIII).

$tan(ϕdrv)=−bωdrvk−mωdrv2$ (IX)

$sin(ϕdrv)=−bωdrv(bωdrv)2+(k−mωdrv2)2$ (X)

$cos(ϕdrv)=(k−mωdrv2)(bωdrv)2+(k−mωdrv2)2$ (XI)

Substitute $−bωdrv(bωdrv)2+(k−mωdrv2)2$ for $sin(ϕdrv)$ and $(k−mωdrv2)(bωdrv)2+(k−mωdrv2)2$ for $cos(ϕdrv)$ in equation (VII) to find $Adrv$ .

$Fmax=Adrv[(k−mωdrv2)[(k−mωdrv2)(bωdrv)2+(k−mωdrv2)2]−bωdrv[−bωdrv(bωdrv)2+(k−mωdrv2)2]]=Adrv(bωdrv)2+m2(ω2−ωdrv2)Adrv=Fmax(bωdrv)2+m2(ω2−ωdrv2)$

Thus, the solution is satisfying both side of the equation.

Answer 6

Chapter 16, Problem 82PQ

To determine

To prove the relation for system.

Expert Solution & Answer

Answer to Problem 82PQ

The solution to the equation is $Fmaxcos(ωdrvt)=m(d2y/dt2)+b(dy/dt)+ky_$ .

Explanation of Solution

Write the expression for the given solution of the equation.

$y(t)=Adrvcos(ωdrvt+ϕdrv)$ (I)

Here, $y(t)$ is the instantaneous displacement, $Adrv$ is the amplitude, $ωdrv$ is the angular frequency, $t$ is the time and $ϕdrv$ is the phase angle.

Write the expression for the equation.

$Fmaxcos(ωdrvt)=m(d2y/dt2)+b(dy/dt)+ky$ (II)

Here, $Fmax$ is the maximum force, $(d2y/dt2)$ is the second derivative and $(dy/dt)$ is the first derivative.

Write the expression for the angular frequency.

$ωdrv=km$ (III)

Here, $k$ is wave vector and $m$ is mass.

Conclusion:

Taking first and second derivative from equation (I).

$dydt=−ωdrvAdrvsin(ωdrvt+ϕdrv)$ (IV)

$d2ydt2=−ωdrv2Adrvcos(ωdrvt+ϕdrv)$ (V)

Substitute $−ωdrvAdrvsin(ωdrvt+ϕdrv)$ for $dydt$ , $−ωdrv2Adrvcos(ωdrvt+ϕdrv)$ for $d2ydt2$ and $Adrvcos(ωdrvt+ϕdrv)$ for $y(t)$ in Equation (II) to find $Fmaxcos(ωdrvt)$ .

$Fmaxcos(ωdrvt)=(m[−ωdrv2Adrvcos(ωdrvt+ϕdrv)]+b[−ωdrvAdrvsin(ωdrvt+ϕdrv)]+k[Adrvcos(ωdrvt+ϕdrv)])$ (VI)

Using trigonometric identity and comparing both side the components of sine and cosine terms.

$Fmax=Adrv[(k−mωdrv2)cos(ϕdrv)−bωdrvsinϕdrv]$ (VII)

$0=Adrv[−(k−mωdrv2)sin(ϕdrv)−bωdrvcosϕdrv]$ (VIII)

Write the expression for the trigonometric identities by using equation (VIII).

$tan(ϕdrv)=−bωdrvk−mωdrv2$ (IX)

$sin(ϕdrv)=−bωdrv(bωdrv)2+(k−mωdrv2)2$ (X)

$cos(ϕdrv)=(k−mωdrv2)(bωdrv)2+(k−mωdrv2)2$ (XI)

Substitute $−bωdrv(bωdrv)2+(k−mωdrv2)2$ for $sin(ϕdrv)$ and $(k−mωdrv2)(bωdrv)2+(k−mωdrv2)2$ for $cos(ϕdrv)$ in equation (VII) to find $Adrv$ .

$Fmax=Adrv[(k−mωdrv2)[(k−mωdrv2)(bωdrv)2+(k−mωdrv2)2]−bωdrv[−bωdrv(bωdrv)2+(k−mωdrv2)2]]=Adrv(bωdrv)2+m2(ω2−ωdrv2)Adrv=Fmax(bωdrv)2+m2(ω2−ωdrv2)$

Thus, the solution is satisfying both side of the equation.

Answer 7

Chapter 16, Problem 82PQ

To determine

To prove the relation for system.

Answer 8

Expert Solution & Answer

Answer to Problem 82PQ

The solution to the equation is $Fmaxcos(ωdrvt)=m(d2y/dt2)+b(dy/dt)+ky_$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Write the expression for the given solution of the equation.

$y(t)=Adrvcos(ωdrvt+ϕdrv)$ (I)

Here, $y(t)$ is the instantaneous displacement, $Adrv$ is the amplitude, $ωdrv$ is the angular frequency, $t$ is the time and $ϕdrv$ is the phase angle.

Write the expression for the equation.

$Fmaxcos(ωdrvt)=m(d2y/dt2)+b(dy/dt)+ky$ (II)

Here, $Fmax$ is the maximum force, $(d2y/dt2)$ is the second derivative and $(dy/dt)$ is the first derivative.

Write the expression for the angular frequency.

$ωdrv=km$ (III)

Here, $k$ is wave vector and $m$ is mass.

Conclusion:

Taking first and second derivative from equation (I).

$dydt=−ωdrvAdrvsin(ωdrvt+ϕdrv)$ (IV)

$d2ydt2=−ωdrv2Adrvcos(ωdrvt+ϕdrv)$ (V)

Substitute $−ωdrvAdrvsin(ωdrvt+ϕdrv)$ for $dydt$ , $−ωdrv2Adrvcos(ωdrvt+ϕdrv)$ for $d2ydt2$ and $Adrvcos(ωdrvt+ϕdrv)$ for $y(t)$ in Equation (II) to find $Fmaxcos(ωdrvt)$ .