Question
Jan Villaroel
Topic: Physics Posted 1 year ago
A pendulum is released from an angle of 30 degrees from the vertical. Determine the period, frequency, and maximum speed of the pendulum bob.

A pendulum is released from an angle of 30 degrees from the vertical. Determine the period, frequency, and maximum speed of the pendulum bob.


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Koala Tutor
Oct 3, 2023
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The period of a simple pendulum is given by the formula:

T = 2π√(L / g)

where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.

Since the length of the pendulum is not given, we can assume it to be a variable, L.

The maximum speed of the pendulum bob occurs when it is at the bottom of its swing, at an angle of 30 degrees from the vertical. At this point, all of the potential energy is converted to kinetic energy. The formula for maximum speed is:

v = √(2gL(1 - cosθ))

where v is the maximum speed, g is the acceleration due to gravity, L is the length of the pendulum, and θ is the angle from the vertical.

In this case, g is a constant, so we can denote it as "g".

Given: θ = 30 degrees

Substituting this value into the formula for maximum speed:

v = √(2gL(1 - cos(30)))

To find the maximum speed, we need to know the length of the pendulum.

The frequency of the pendulum is the reciprocal of the period:

f = 1 / T

where f is the frequency and T is the period.

Therefore, to find the frequency, we need to find the period first.

We can rearrange the formula for the period to solve for L:

T = 2π√(L / g)
T^2 = 4π^2(L / g)
(L / g) = T^2 / (4π^2)
L = T^2 * g / (4π^2)

We are given that the angle is 30 degrees, which is equivalent to π/6 radians.

Substituting the value of L into the formula for maximum speed:

v = √(2g(T^2 * g / (4π^2))(1 - cos(π / 6)))

Simplifying:

v = √((g^2T^2) / (2π^2))(1 - cos(π / 6))
v = √(g^2T^2 / (2π^2))(1 - √3/2)
v = √(g^2T^2 / (2π^2))(2 - √3)
v = (gT / (2π))(2 - √3)

To find the period, we substitute the value of L back into the formula for the period:

T = 2π√((T^2 * g / (4π^2)) / g)
T = 2π√(T^2 / (4π^2))
T = 2πT / (2π)
T = T

Therefore, the period (T) is the same as the initial value we assumed.

The frequency is given by:

f = 1 / T
f = 1 / (T / (2π))
f = 2π / T
f = 2π / T

The maximum speed (v) is given by:

v = (gT / (2π))(2 - √3)

To find the numerical values for T, f, and v, we need to know the length of the pendulum (L) and the acceleration due to gravity (g).

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