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Q49 P

Expert-verifiedFound in: Page 438

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Question: The angle of the pendulum in Figure is given by ** **. If at t = 0 ** **, ** ** and ** **, **

**what is the phase constant****,****what is the maximum angle****?**

**(Hint: Don’t confuse the rate at which changes with the ** ** of the SHM.)**

**Answer**

** **

- The phase constant is 0.845 rad
- The maximum angle is 0.0602 rad

** **

** **

- The angle of the pendulum is
- At , and

** ****The oscillations of the simple pendulum can be defined by the equation of simple harmonic motion. The simple harmonic motion is the motion in which the acceleration of the oscillating object is directly proportional to the displacement. The force caused by the acceleration is called restoring force. This restoring force is always directed towards the mean position.**

** **

**Compare the given equation with the equation of displacement of the particle in simple harmonic motion.**

** **

Formulae:

** **The phase constant :

The expression for the displacement of the particle in simple harmonic motion is

Here, x (t) is the displacement, x_{m} is amplitude, angular velocity, t is time, is phase difference.

For angular displacement, replace x by , then

…(i)

The expression for velocity of the particle in simple harmonic motion is

For angular motion, replace x by , then

…(ii)

Divide equation (ii) by equation (i)

Therefore, the phase constant is 0.845 rad .

** **

For t =0, equation (i) becomes as

The maximum angle is 0.0602 rad

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