Question

A simple pendulum 2 m long swings through a maximum angle of 30 ∘ with the vertical. part a calculate its period assuming a small amplitude.

Answer 1

The time period of the pendulum performing the oscillation of small amplitude is $\boxed{2.837\,{\text{s}}}$.

Further Explanation:

Given:

The length of the simple pendulum is $2\,{\text{m}}$.

The maximum angle of the pendulum’s oscillation is $30^\circ$.

Concept:

If the amplitude of oscillation of the simple pendulum is very small, then the oscillation of the pendulum is termed as the simple harmonic.

The time period of a given pendulum is independent of the mass of the bob and amplitude of oscillation. It depends on the length and the acceleration due to gravity of the Earth.

The time period of a pendulum exhibiting simple harmonic motion is given as:

$\boxed{T = 2\pi \sqrt {\frac{L}{g}} }$                         …… (1)

Here, $T$ is the time period of oscillation of the pendulum, $L$ is the length of the pendulum and $g$ is the acceleration due to gravity of the Earth.

The value of acceleration due to gravity on the Earth is $9.8\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}$.

Substitute the values of length and acceleration due to gravity in above expression.

$\begin{aligned}T&= 2\pi \sqrt {\frac{2}{{9.8}}}\\&= 6.28 \times 0.451\,{\text{s}} \\&=2.837\,{\text{s}} \\ \end{aligned}$

Thus, the time period of the pendulum performing the oscillation of small amplitude is $\boxed{2.837\,{\text{s}}}$.

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Answer Details:

Grade: Middle School

Subject: Physics

Chapter: Simple Harmonic Motion

Keywords:

Simple pendulum, harmonic motion, maximum angle, 30 degree, with vertical, oscillation, time period, small amplitude.

Answer 2

Length of the pendulum (l) = 2 m

Acceleration due to gravity = g = 9.8 m/s^2

For small amplitude, the pendulum will undergo simple harmonic motion.

Hence, the time period of the pendulum for small amplitude = $2\pi \sqrt{\frac{l}{g} }$

Now, plug the values of l and g

T = $2\pi \sqrt{\frac{2}{9.8} }$

T = 2 × 3.14 × 0.451

T = 2.83 seconds

Hence, the time period of the pendulum for small amplitude = 2.83 s