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 46P

(a)

To determine

The position of x at a moment 84.45s later.

(a)

Expert Solution
Check Mark

Answer to Problem 46P

The position of x at a moment 84.45s later is 15.8cm_

Explanation of Solution

The general expression for the position of the object is given by,

    x=Acosωt        (I)

Here, x is the position of the object, A is the amplitude, ω is the angular frequency, t is the time.

Write the expression for the force due to a spring.

    F=kx        (II)

Here, F is the force acting due to spring, k is the spring constant, x is the compression distance.

Write the expression for the force acting on an object.

    F=mg        (III)

Here, g is the acceleration due to gravity, m is the mass of the object.

Equate equation (II) and (III) and solve for k.

    kx=mgk=mgk        (IV)

Take the x axis as pointing downward,

Write the expression for the angular frequency.

    ω=km        (V)

Conclusion:

Substitute 450g for m and 35.0cm for x in equation (IV) to find k.

    k=(450g×1kg103g)9.8m/s235.0cm×1m100cm=12.6N/m

Substitute 12.6N/m for k, 450g for m in equation (V) to find ω.

    ω=12.6N/m450g×1kg103g=5.29rad/s

Substitute 18.0cm for A, 5.29 for ω and 84.4s for t in equation (I) to find x.

    x=(18.0cm)cos(5.29rad/s)(84.4s)=15.8cm

Therefore, The position of x at a moment 84.45s later is 15.8cm_

(b)

To determine

The distance traveled by the vibrating object in part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 46P

The distance traveled by the vibrating object in part (a) is 51.1m_.

Explanation of Solution

The distance traveled is equal to the number of oscillations multiplied by the distance traveled in a single time period.

The time period is given by,

    T=2πω        (VI)

Here, T is the time period.

The number of oscillations made by the spring is given by,

    n=tT        (VI)

Here, t is the total time of the oscillations.

The distance travelled by the mass in one complete oscillation is 4A.

The total distance traveled in 84.4s is,

    d=4An        (VII)

Conclusion:

Substitute 5.29rad for ω in equation (VI) to find T.

    T=2π5.29=1.1874s

Substitute 1.1874s for T, 84.4s for t in equation (VI) to find n.

    n=84.4s1.1874s=71.079

Substitute e 0.18m for A and 71.079 for n in equation (

  d=4(0.18m)(71.079)=51.1769m=51.2m

Therefore, The distance traveled by the vibrating object in part (a) is 51.2m_.

(c)

To determine

The position of the object 84.4s later when an object of mass 440g is bung on it at rest.

(c)

Expert Solution
Check Mark

Answer to Problem 46P

The position of the object 84.4s later when an object of mass 440g is bung on it at rest is 15.9cm_

Explanation of Solution

Substitute 0.440g for m, 9.80m/s2 for g and 35.5cm for x in equation (IV) to find k.

        k=(440g×1kg103g)9.8m/s235.5cm×1m100cm=12.1N/m

Substitute 12.1N/m for k, 440g for m in equation (V) to find ω.

    ω=12.1N/m440g×1kg103g=5.244rad/s

Substitute 18.0cm for A, 5.244 for ω and 84.4s for t in equation (I) to find x.

    x=(18.0cm)cos(5.24rad/s)(84.4s)=15.9cm

Therefore, the position of x at a moment 84.45s later is 15.9cm_

Conclusion:

Therefore, the position of the object 84.4s later when an object of mass 440g is bung on it at rest is 15.9cm_

(d)

To determine

The distance traveled by the object in part (c)

(d)

Expert Solution
Check Mark

Answer to Problem 46P

The distance traveled by the object in part (c) is 50.8cm_.

Explanation of Solution

The distance traveled is equal to the number of oscillations multiplied by the distance traveled in a single time period.

The time period is given by,

    T=2πω

The number of oscillations made by the spring is given by,

    n=tT

The distance travelled by the mass in one complete oscillation is 4A.

The total distance traveled in 84.4s is,

    d=4An

Conclusion:

Substitute 5.244rad for ω in equation (VI) to find T.

    T=2π5.24=1.199s

Substitute 1.199s for T, 84.4s for t in equation (VI) to find n.

    n=84.4s1.199s=70.39

Substitute e 0.18m for A and 71.079 for n in equation (

  d=4(0.18m)(70.39)=50.8m

Therefore, the distance traveled by the vibrating object in part (c) is 50.8m_.

(e)

To determine

The reason for the different answers to part (a) and (c) when the initial data in parts (a) and (c) are so similar and the answers to parts (b) and (d) are relatively close.

(e)

Expert Solution
Check Mark

Answer to Problem 46P

The answers in parts a and b are different because of the difference in angular velocities.

Explanation of Solution

Diverging patterns of oscillations which starts out in phase but becoming completely out of phase changes the answers in part (a) and (b). Since the angular velocity is changing it will lead to the change in the position of the phase. But this phase difference is higher when compared to the angular velocity difference as it is multiplied by time also. Thus it yields larger change in the position.

For parts c and d, the distance traveled depends only on the angular velocity such that their difference is not so large.

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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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