A pendulum of length l metres takes t seconds to complete one full oscillation. The formula for T is:
T = 2π √(l/g)
Where g m/s2 is the acceleration due to gravity.
a) Find T if l = 5 and g = 10
b) Rearrange the formula to make l the subject of the formula
c) How long is a pendulum which takes 3 seconds for one oscillation, if g = 10?
(a) just plug in your numbers
(b) T = 2π √(l/g)
T / 2π = √(l/g)
(T / 2π)^2 = (l/g)
now finish it off and (c) is a cinch.
a) To find T when l = 5 and g = 10, substitute the given values into the formula:
T = 2π √(l/g)
T = 2π √(5/10)
T = 2π √(1/2)
T = 2π / √2
Approximately, T = 4.442 seconds.
b) To rearrange the formula to make l the subject, start with the given formula:
T = 2π √(l/g)
To isolate l, square both sides of the equation:
T^2 = (2π)^2 * (l/g)
Divide both sides by (2π)^2 * g:
T^2 / ((2π)^2 * g) = l
Therefore, the rearranged formula is:
l = T^2 / ((2π)^2 * g)
c) To find the length of the pendulum which takes 3 seconds for one oscillation, substitute T = 3 and g = 10 into the formula:
l = T^2 / ((2π)^2 * g)
l = 3^2 / ((2π)^2 * 10)
l = 9 / ((2π)^2 * 10)
l = 9 / (4π^2 * 10)
l = 9 / (12.566 * 10)
l = 9 / 125.66
Approximately, l = 0.0716 meters.
a) To find T when l = 5 and g = 10, we can substitute these values into the formula:
T = 2π √(l/g)
T = 2π √(5/10)
T = 2π √(0.5)
Now we can simplify the square root:
T = 2π √(0.5)
T ≈ 2π * 0.707
Using an approximation for π (3.14), we can calculate the value of T:
T ≈ 2 * 3.14 * 0.707
T ≈ 4.44 seconds
So, when l = 5 and g = 10, the pendulum takes approximately 4.44 seconds for one full oscillation.
b) To rearrange the formula to make l the subject, we can start by squaring both sides of the equation:
T^2 = (2π √(l/g))^2
T^2 = 4π^2 (l/g)
Now, we can multiply both sides of the equation by (g/4π^2):
(T^2) * (g/4π^2) = l
Therefore, the formula with l as the subject is:
l = (T^2) * (g/4π^2)
c) To find the length of a pendulum that takes 3 seconds for one oscillation when g = 10, we can use the rearranged formula:
l = (T^2) * (g/4π^2)
Substituting T = 3 and g = 10 into the formula:
l = (3^2) * (10/4π^2)
l = 9 * (10/4π^2)
l ≈ 90/(4π^2)
Using an approximation for π (3.14), we can calculate the value:
l ≈ 90/(4 * 3.14^2)
l ≈ 90/(4 * 9.86)
l ≈ 90/39.44
l ≈ 2.28 meters
Therefore, a pendulum that takes 3 seconds for one oscillation, with g = 10, has a length of approximately 2.28 meters.
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