Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card)
Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card)
5th Edition
ISBN: 9781305586871
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 12, Problem 49P

(a)

To determine

The equation of the motion of the particle.

(a)

Expert Solution
Check Mark

Answer to Problem 49P

The motion equation of the motion of the particle is x=2cos(10t+π2) .

Explanation of Solution

Given information:

Mass of the particle is 0.50kg , the force constant of the spring is 50.0N/m and the maximum speed of the particle at time t=0 is 20.0m/s .

Formula to calculate the angular frequency is,

ω=km

The general equation of the particle’s motion is,

x=Acos(ωt+ϕ) (I)

Differentiate the above equation with respect to time.

dxdt=ddt(Acos(ωt+ϕ))v=Aωsin(ωt+ϕ)

From the given condition, At t=0 , v=vmax

Substitute these values in the above equation.

vmax=Aωsin(ω×0+ϕ)vmax=Aωsinϕ

The maximum value of sinϕ is 1. Therefore,

sinϕ=1ϕ=π2

Therefore,

vmax=AωA=vmaxω

Substitute km for ω in the above equation.

A=vmaxmk

Substitute vmaxmk for A , km for ω and π2 for ϕ in equation (I).

x=vmaxmkcos(kmt+π2)

Substitute 20.0m/s for vmax , 50.0N/m for k and 0.50kg for m in the above equation.

x=20.0m/s×0.50kg50.0N/mcos(50.0N/m0.50kgt+π2)=2cos(10t+π2)

Here, x in meter and t in seconds.

Conclusion:

Therefore, the motion equation of the motion of the particle is x=2cos(10t+π2) .

(b)

To determine

The position where the potential energy is the three times the kinetic energy.

(b)

Expert Solution
Check Mark

Answer to Problem 49P

The position where the potential energy is the three times the kinetic energy is ±1.73m .

Explanation of Solution

Given information:

Mass of the particle is 0.50kg , the force constant of the spring is 50.0N/m and the maximum speed of the particle at time t=0 is 20.0m/s .

Formula to calculate the maximum energy stored in the spring is,

Umax=12kA2

Formula to calculate the potential energy at any position is,

U=12kx2

Formula to calculate the kinetic energy at any position is,

K=12mv2

From the given condition,

12mv2=13(12kx2)

From the conservation of energy,

U+K=Umax12kx2+12mv2=12kA2

Substitute 13(12kx2) for 12mv2 in the above equation.

12kx2+13(12kx2)=12kA243x2=A2x=±34A

Substitute 2.0m for A in the above equation.

x=±34×2.0m=±1.73m

Conclusion:

Therefore, the position where the potential energy is the three times the kinetic energy is ±1.73m .

(c)

To determine

The minimum time interval required for the particle to move from x=0 to x=1.0m .

(c)

Expert Solution
Check Mark

Answer to Problem 49P

The minimum time interval required for the particle to move from x=0 to x=1.0m is 105ms .

Explanation of Solution

Given information:

Mass of the particle is 0.50kg , the force constant of the spring is 50.0N/m and the maximum speed of the particle at time t=0 is 20.0m/s .

The position of the particle is given by,

x=2cos(10t+π2)

So, the particle will be at x=0 when,

10t+π2=π2,3π2,5π2...10t=0,π,2π,3π... (II)

Initially, at t=0 the particle is at origin and moving to the left. So, next time it will be at origin when 10t=π that is moving to the right.

So, the particle will be at x=1.0m when,

10t+π2=3π2+π310t=11π6π2=4π3 (III)

Therefore, the minimum time interval required for the particle to move from x=0 to x=1.0m is,

10Δt=4π3πΔt=π3×10=0.105s×103ms1s=105ms

Conclusion:

Therefore, the minimum time interval required for the particle to move from x=0 to x=1.0m is 105ms .

(d)

To determine

The length of a simple pendulum with same period.

(d)

Expert Solution
Check Mark

Answer to Problem 49P

The length of a simple pendulum with same period is 0.098m .

Explanation of Solution

Given information:

Mass of the particle is 0.50kg , the force constant of the spring is 50.0N/m and the maximum speed of the particle at time t=0 is 20.0m/s .

Formula to calculate the length of the pendulum is,

ω=gLL=gω2

L is the length of the pendulum.

g is te acceleration due to gravity.

Substitute 9.8m/s2 for g and 10rad/s for ω in the above equation.

L=9.8m/s2(10rad/s)2=0.098m

Conclusion:

Therefore, the length of a simple pendulum with same period is 0.098m .

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Chapter 12 Solutions

Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card)

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