Physics: Principles with Applications
Physics: Principles with Applications
6th Edition
ISBN: 9780130606204
Author: Douglas C. Giancoli
Publisher: Prentice Hall
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Chapter 11, Problem 32P

(a)

To determine

To Calculate: The frequency of simple pendulum.

(a)

Expert Solution
Check Mark

Answer to Problem 32P

The frequency at which it vibrates is 0.572 Hz .

Explanation of Solution

Given:

Length of simple pendulum  l = 0.76 m

Formula used:

Frequency of the pendulum is obtained by the following formula,

   f = 12π×gl

Here,

  g is the acceleration due to gravity.

  l is the length of simple pendulum.

Calculation:

Substitute the given values in the above equation,

  Frequency f = 12π×9.80.76 = 0.572 Hz

Conclusion:

Thus, the frequency of pendulum at which it vibrates is 0.572 Hz .

(b)

To determine

To Calculate: The speed of pendulum bob.

(b)

Expert Solution
Check Mark

Answer to Problem 32P

The speed of pendulum bob is 0.571 m/s .

Explanation of Solution

Given:

Length of pendulum  l = 0.76 m

Angle θ = 12°

Value of kinetic energy at top (at release point), KE (top) = 0 

  KE (bottom) = 12×m×vmax2  (at lowest point KE is maximum)

Value of potential energy at top PE (top) = m×g×(L - Lcosθ)

Value of potential energy at bottom PE (bottom) = 0

Formula used:

The conservation of energy can be used and related to the lowest point to the release point of pendulum, speed of the pendulum bob is determined by the following formula,

  E(top) = E(bottom)

It can be written as

  KE(top) + PE(top) = KE(bottom) + PE(bottom)

Here KE represent the kinetic energy of the bob and  PE represent the potential energy of the bob.

Calculation:

Substitute the given values,

  KE(top) + PE(top) = KE(bottom) + PE(bottom)0 + mg (L - Lcosθ) = 12×m×vmax2 + 0vmax2  = 2×g×(L - Lcosθ) = 2×9.8×0.76×(1 - cos12°)vmax = 0.571 m/s

The speed of pendulum bob is 0.571 m/s .

Conclusion:

Thus, the speed of pendulum bob is 0.571 m/s .

(c)

To determine

To Calculate: The total energy stored in oscillation.

(c)

Expert Solution
Check Mark

Answer to Problem 32P

The total energy stored in this oscillation is 5.95×10-2 J .

Explanation of Solution

Given:

Maximum velocity vmax = 0.571 m/s

Mass m = 0.365 kg

Formula used:

The total energy can be found from the KE at the bottom of motion by the following formula,

  Etotal = 12×m×vmax2

Here,

  m is mass of pendulum.

  vmax is maximum speed of bob.

Calculation:

Substitute the given values in the above equation,

  Etotal = 12×0.365×0.5712Etotal= 5.95×10-2 J

Conclusion:

Thus, the total energy stored in this oscillation is 5.95×10-2 J .

Chapter 11 Solutions

Physics: Principles with Applications

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