- #1

- 202

- 0

[itex]T = 2 \pi \sqrt{\frac{l}{g}}[/itex]

Thanks in advance.

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- Thread starter Air
- Start date

- #1

- 202

- 0

[itex]T = 2 \pi \sqrt{\frac{l}{g}}[/itex]

Thanks in advance.

- #2

- 47

- 0

T denotes the period of oscillation,

l the length of the string on which the mass hangs.

Notice that the period doesn't depend on how "wide" the mass swings.

- #3

- 147

- 0

l : m/s

g: m/s^2

(m/s) / (m/s^2) : s

- #4

- 202

- 0

[itex]P = \frac{1}{T}[/itex]

So to calculate time of oscillation, [itex]T[/itex] would need to be made subject.

- #5

- 147

- 0

frequence : 1/s

period : s

period : s

- #6

HallsofIvy

Science Advisor

Homework Helper

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- 967

No, a "period" is a certain length of time and so has the same units as T (whatever "T" means here). The formula you give is for "frequency"- the number of times the pendulum swings during 1 second and so has units of "1/sec".

[itex]P = \frac{1}{T}[/itex]

So to calculate time of oscillation, [itex]T[/itex] would need to be made subject.

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