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Andrews [41]
3 years ago
14

The period of a simple pendulum, defined as the time necessary for one complete oscillation, is measured in time units and is gi

ven by
T = 2π×square root of l/g

where ℓ is the length of the pendulum and g is the acceleration due to gravity, in units of length divided by time squared. Show that this equation is dimensionally consistent.
Physics
1 answer:
Oksana_A [137]3 years ago
7 0

The task is to show that the right side of the equation has units of [Time], just like the left side has.

The right side of the equation is . . . 2 π √(L/G) .

We can completely ignore the  2π since it has no units at all, so it has no effect on the units of the right side of the equation.  Now the task is simply to find the units of  √(L/G) .

L . . . meters

G . . . meters/sec²

(L/G) = (meters) / (meters/sec²)

(L/G) = (meters) · (sec²/meters)

(L/G) = (meters · sec²) / (meters)

(L/G) = sec²

So  √(L/G) = seconds = [Time]

THAT's what we were hoping to prove, and we did it !

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Explanation:

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6 0
4 years ago
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3 years ago
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Answer:

g = 11.2 m/s²

Explanation:

First, we will calculate the time period of the pendulum:

T = \frac{t}{n}

where,

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T = \frac{135\ s}{98}

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Now, we utilize the second formula for the time period of the simple pendulum, given as follows:

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3 years ago
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Answer

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