Question 174: Oscillations

Question

The length of a simple pendulum executing simple harmonic motion is increased by $$21\% $$. The percentage increase in the time period of the pendulum of increased length is

Accepted Answer

The period of a simple pendulum is given by:
$$ T = 2\pi \sqrt{\frac{L}{g}} $$
where:
T is the period,
L is the length of the pendulum, and
g is the acceleration due to gravity.
Since g is constant, we see that the period T is proportional to the square root of the length L.
If L is increased by 21%, the new length L' is L + 21%L = 1.21L. The new period T' is then:
$$ T' = 2\pi \sqrt{\frac{L'}{g}} = 2\pi \sqrt{\frac{1.21L}{g}} = \sqrt{1.21}T \approx 1.1T $$
The percentage increase in the time period is then:
$$ \frac{T' - T}{T} 	imes 100\% = (\sqrt{1.21} - 1) 	imes 100\% \approx 10\% $$
Therefore, the percentage increase in the time period of the pendulum of increased length is approximately 10%.

Answer

$$11\% $$

Answer

$$21\% $$

Answer

$$42\% $$

Answer

$$10\% $$

Oscillations

The length of a simple pendulum executing simple harmonic motion is increased by $21\% $. The percentage increase in the time period of the pendulum of increased length is

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