Find the speed of a pendulum bob at the bottom of its swing if its initial displacement is 5.34o and its period is 2.534 s.

P = 2 pi sqrt(L/g)

Solve for L
L - Lcos5.34 = h
now
mgh = 1/2mv^2
and the m's cancel, solve for v

To find the speed of a pendulum bob at the bottom of its swing, we can use the formula for the period of a simple pendulum:

T = 2 * π * √(L/g),

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

However, in this case, we are given the initial displacement (5.34°) and the period (2.534 s), so we need to use a slightly different approach.

First, we need to find the length of the pendulum. We can use the following relationship between the period T and the length L:

T = 2 * π * √(L/g).

Rearranging the equation to solve for L, we get:

L = (T^2 * g) / (4 * π^2).

Plugging in the values for T (2.534 s) and g (9.8 m/s^2), we can find the length L of the pendulum.

L = (2.534^2 * 9.8) / (4 * π^2) = 1.007 m (rounded to three decimal places).

Now that we have the length of the pendulum, we can find the speed of the bob at the bottom of its swing.

At the bottom of its swing, the potential energy of the bob is converted into kinetic energy. The potential energy at the top of the swing is given by:

PE = m * g * h,

where m is the mass of the bob, g is the acceleration due to gravity, and h is the height.

Since the bob is at the bottom of its swing, the height h is equal to the length L of the pendulum.

The kinetic energy at the bottom of the swing is given by:

KE = (1/2) * m * v^2,

where v is the speed of the bob at the bottom of its swing.

According to the law of conservation of energy, the potential energy at the top equals the kinetic energy at the bottom:

PE = KE,

m * g * L = (1/2) * m * v^2,

Canceling the mass m on both sides and solving for v, we get:

v = √(2 * g * L).

Plugging in the values for g (9.8 m/s^2) and L (1.007 m), we can find the speed v of the bob at the bottom of its swing:

v = √(2 * 9.8 * 1.007) = 4.484 m/s (rounded to three decimal places).

Therefore, the speed of the pendulum bob at the bottom of its swing is approximately 4.484 m/s.