Question

A 100 g mass on a 1.0 m long string is pulled 8.0o
to one side and released. How long does
it take for the pendulum to reach 4.0o on the opposite side?

Answer 1

Answer on Question #45369, Physics, Mechanics | Kinematics | Dynamics

A $100\mathrm{g}$ mass on a $1.0\mathrm{m}$ long string is pulled $8.0{}^{\circ}$ to one side and released. How long does it take for the pendulum to reach $4.0{}^{\circ}$ on the opposite side?

Solution:

The motion of a simple pendulum is like simple harmonic motion.

Swing time "there and back" (period)

For small amplitudes, the period (swing time "there and back") of such a pendulum can be approximated by:

$$T = 2 \pi \sqrt {\frac {L}{g}}$$

where $L$ is the length of the pendulum and $g = 9.81 \, \text{m/s}^2$ .

Thus, in our case

$$T = 2 \cdot 3. 1 4 \sqrt {\frac {1 . 0}{9 . 8 1}} = 2 \mathrm {s}$$

The time to reach equilibrium position from $8.0{}^{\circ}$ is

$$t _ {0} = \frac {T}{4} = 0. 5 \mathrm {s}$$

The motion of a simple pendulum is like simple harmonic motion in that the equation for the angular displacement is

$$\theta = \theta_ {m a x} \sin \sqrt {\frac {g}{L}} t$$

The time to reach $4.0{}^{\circ}$ from equilibrium position is

$$t _ {1} = \sqrt {\frac {L}{g}} \sin^ {- 1} \left(\frac {\theta}{\theta_ {m a x}}\right) = \sqrt {\frac {1 . 0}{9 . 8 1}} \cdot \sin^ {- 1} \left(\frac {4}{8}\right) = 0. 1 6 7 \approx 0. 1 7 \mathrm {s}.$$

Thus, the time to reach $4.0{}^{\circ}$ on the opposite side is

$$t = t _ {0} + t _ {1} = 0. 5 + 0. 1 7 = 0. 6 7 \mathrm {s}.$$

Answer: $t = 0.67$ s.

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