What is the period of a pendulum near Earth's surface that is 130 cm long?

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What is the period of a pendulum near Earth's surface that is 130 cm long?

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Answer 1 · 2017-05-25T11:18:27

$T \approx 2.2876585681 s$ $T~~2.2876585681s$

Explanation:

Since you were not given the degrees of amplitude or radians of amplitude, we assume that the degrees of amplitude are less than $20^{\circ}$ $20^@$ or $20^{\circ}$ $20^@$ and we also assume that this is a simple pendulum

Apply the equation

$T \approx 2 π \sqrt{\frac{L}{g}} (1)$ $T ~~ 2pisqrt(L/g)" "(1)$

And for accurate

$T = \frac{2 π}{amg (1, cos (\frac{π \cdot degrees}{360^{\circ}}))} \times \sqrt{\frac{L}{g}} (2)$ $T = (2pi)/("amg"(1,cos((pi * "degrees")/360^@)))xxsqrt(L/g)" "(2)$

Where $g$ $g$ is acceleration due to gravity that is 9.80665m/s

$amg(arithmetic geometric mean)$ $"amg(arithmetic geometric mean)"$ is an operation

T is the time period

Time period is the time required for one oscillation

Where L is length

But we would use equation one

$T \approx 2 π \sqrt{\frac{1.3m}{9.80665m/s}}$ $T ~~ 2pisqrt("1.3m"/"9.80665m/s")$

The unit for length here is m

$T \approx 2.2876585681 s$ $T~~2.2876585681s$

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What is the period of a pendulum near Earth's surface that is 130 cm long?

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