Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 14, Problem 92P

(a)

To determine

The time period of oscillating system.

(a)

Expert Solution
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Explanation of Solution

Introduction:

Time period of oscillating system is the time required by system to complete one oscillation.

The free body diagram of system is drawn below.

  Physics for Scientists and Engineers, Chapter 14, Problem 92P , additional homework tip  1

Here, Mg is the weight of bob, kx is the spring force, L is the length of pendulum, T is the tension in string and θ is the angle of string with the vertical.

Write the expression for time period of the pendulum.

  Tp=2πω …… (1)

Here, Tp is time period and ω is angular frequency of pendulum.

Write the expression for resultant force in horizontal direction.

  Fx=0

Substitute kxTsinθMax for Fx in above expression.

  kxTsinθMax=0 …… (2)

Here, k is the spring constant, x is the extension in length, T is the tension, M is the mass of bob, ax is the acceleration and θ is the angle with the vertical.

Write the expression for resultant force in horizontal direction.

  Fy=0

Substitute TcosθMg for Fy in above expression.

  TcosθMg=0

Here, g is the gravitational acceleration.

Rearrange the above expression in term of T .

  T=Mgcosθ

Substitute Mgcosθ for T in equation (1) and simplify.

  kxMgtanθMax=0

Substitute Lθ for x and Ld2θdt2 for ax in above expression.

  kLθMgtanθMLd2θdt2=0

Rearrange the above expression in term of d2θdt2 for θ1 .

  d2θdt2=(kM+gL)θ

Compare the above expression with the equation of differential equation of S.H.M. d2θ/dt2=ω2θ .

The value of angular frequency is ω=kM+gL

Substitute kM+gL for ω in equation (1).

  Tp=2πkM+gL …… (3)

Conclusion:

Thus, the time period of oscillating system is Tp=2πkM+gL .

(b)

To determine

The value of force constant.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

The mass of bob is 1.00kg .

The first time period of the pendulum is 2.00s .

The second time period of the pendulum is 1.00s .

Formula:

The free body diagram of system is drawn below.

  Physics for Scientists and Engineers, Chapter 14, Problem 92P , additional homework tip  2

Here, Mg is the weight of bob, kx is the spring force, L is the length of pendulum, T is the tension in string and θ is the angle of string with the vertical.

Write the expression for time period of the pendulum.

  Tp=2πω …… (1)

Here, Tp is time period and ω is angular frequency of pendulum.

Write the expression for resultant force in horizontal direction.

  Fx=0

Substitute kxTsinθMax for Fx in above expression.

  kxTsinθMax=0 …… (2)

Here, k is the spring constant, x is the extension in length, T is the tension, M is the mass of bob, ax is the acceleration and θ is the angle with the vertical.

Write the expression for resultant force in horizontal direction.

  Fy=0

Substitute TcosθMg for Fy in above expression.

  TcosθMg=0

Here, g is the gravitational acceleration.

Rearrange the above expression in term of T .

  T=Mgcosθ

Substitute Mgcosθ for T in equation (1) and simplify.

  kxMgtanθMax=0

Substitute Lθ for x and Ld2θdt2 for ax in above expression.

  kLθMgtanθMLd2θdt2=0

Rearrange the above expression in term of d2θdt2 for θ1 .

  d2θdt2=(kM+gL)θ

Compare the above expression with the equation of differential equation of S.H.M. d2θ/dt2=ω2θ .

The value of angular frequency is ω=kM+gL

Substitute kM+gL for ω in equation (1).

  Tp=2πkM+gL …… (3)

Calculation:

For k=0 , Tp=2.00s :

Substitute 2.00s for Tp , 0 for k , 1.00kg for M in equation (3).

  2.00s=2π01.00kg+gLgL=π2

For Tp=1.00s :

Substitute 1.00s for Tp , π2 for gL and 1.00kg for M in equation (3).

  1.00s=2πk1.00kg+π2k=29.6N/m

Conclusion:

Thus, the value of force constant is 29.6N/m .

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Chapter 14 Solutions

Physics for Scientists and Engineers

Ch. 14 - Prob. 11PCh. 14 - Prob. 12PCh. 14 - Prob. 13PCh. 14 - Prob. 14PCh. 14 - Prob. 15PCh. 14 - Prob. 16PCh. 14 - Prob. 17PCh. 14 - Prob. 18PCh. 14 - Prob. 19PCh. 14 - Prob. 20PCh. 14 - Prob. 21PCh. 14 - Prob. 22PCh. 14 - Prob. 23PCh. 14 - Prob. 24PCh. 14 - Prob. 25PCh. 14 - Prob. 26PCh. 14 - Prob. 27PCh. 14 - Prob. 28PCh. 14 - Prob. 29PCh. 14 - Prob. 30PCh. 14 - Prob. 31PCh. 14 - Prob. 32PCh. 14 - Prob. 33PCh. 14 - Prob. 34PCh. 14 - Prob. 35PCh. 14 - Prob. 36PCh. 14 - Prob. 37PCh. 14 - Prob. 38PCh. 14 - Prob. 39PCh. 14 - Prob. 40PCh. 14 - Prob. 41PCh. 14 - Prob. 42PCh. 14 - Prob. 43PCh. 14 - Prob. 44PCh. 14 - Prob. 45PCh. 14 - Prob. 46PCh. 14 - Prob. 47PCh. 14 - Prob. 48PCh. 14 - Prob. 49PCh. 14 - Prob. 50PCh. 14 - Prob. 51PCh. 14 - Prob. 52PCh. 14 - Prob. 53PCh. 14 - Prob. 54PCh. 14 - Prob. 55PCh. 14 - Prob. 56PCh. 14 - Prob. 57PCh. 14 - Prob. 58PCh. 14 - Prob. 59PCh. 14 - Prob. 60PCh. 14 - Prob. 61PCh. 14 - Prob. 62PCh. 14 - Prob. 63PCh. 14 - Prob. 64PCh. 14 - Prob. 65PCh. 14 - Prob. 66PCh. 14 - Prob. 67PCh. 14 - Prob. 68PCh. 14 - Prob. 69PCh. 14 - Prob. 70PCh. 14 - Prob. 71PCh. 14 - Prob. 72PCh. 14 - Prob. 73PCh. 14 - Prob. 74PCh. 14 - Prob. 75PCh. 14 - Prob. 76PCh. 14 - Prob. 77PCh. 14 - Prob. 78PCh. 14 - Prob. 79PCh. 14 - Prob. 80PCh. 14 - Prob. 81PCh. 14 - Prob. 82PCh. 14 - Prob. 83PCh. 14 - Prob. 84PCh. 14 - Prob. 85PCh. 14 - Prob. 86PCh. 14 - Prob. 87PCh. 14 - Prob. 88PCh. 14 - Prob. 89PCh. 14 - Prob. 90PCh. 14 - Prob. 91PCh. 14 - Prob. 92PCh. 14 - Prob. 93PCh. 14 - Prob. 94PCh. 14 - Prob. 95PCh. 14 - Prob. 96PCh. 14 - Prob. 97PCh. 14 - Prob. 98PCh. 14 - Prob. 99PCh. 14 - Prob. 100PCh. 14 - Prob. 101PCh. 14 - Prob. 103PCh. 14 - Prob. 104PCh. 14 - Prob. 105PCh. 14 - Prob. 106P
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