An image of a physics laboratory with a diverse set of equipment on wooden tables. The focal point is a simple pendulum hanging from a high stand, made of a metal sphere at the end of a thin wire, with an engraved scale behind it showing its angular displacement. The wire length is marked as 0.510 meters. In the background, a hand is slightly pushing the pendulum from its equilibrium position while the other hand is holding a stopwatch.

Question Four

A simple pendulum is used in a physics laboratory experiment to obtain an experimental value for the
gravitational acceleration, g . A student measures the length of the pendulum to be 0.510 meters, displaces
it 10 o
from the equilibrium position, and releases it. Using a stopwatch, the student determines that the
period of the pendulum is 1.44 s. Determine the experimental value of the gravitational acceleration in
two decimal places and the angular speed of the pendulum. (10 marks)

T = 2pi * sqrt L/g,

T^2 = 39.5*L/g = 1.44^2,
39.5 * 0.51/g = 2.074,
20.145/g = 2.074,
g = 9.71 m/s^2.

angular speed will vary in the arc. You could get average angular speed.

Period=2PIsqrt(L/g)
L/g=(period/2PI)^2

g= Length/(period/2PI)^2

No fair using the answer :)

How far does the pendulum drop from the ten degrees?
h = .510 (1 - cos 10) = 0.00775 meters drop
so potential energy at top = m g h = m g (0.00775)

10 degrees * 2 pi radians/360 degrees = .1745 radians
period = 1.44 s
say theta = .1745 sin w t = .1745 sin 2 pi t/T = .1745 sin (2 pi t/1.44)
= .1745 sin (4.36 t)
angular speed = d theta/dt = omega = 0.761 cos (4.36 t)

so max speed at bottom = 0.761 * .510 = .388 m/s
(1/2) m v^2 = m g h which we know is 0.00775 m g
so
g = (.5*.388^2)/0.00775 = 9.72 m/s^2

Oh, gravitational acceleration, you slippery devil. Can't make it easy for us, can you? Alright, let's crunch some numbers here.

First, we need to find the experimental value of the gravitational acceleration. To do that, we can use the period of the pendulum. The formula for the period of a simple pendulum is T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the gravitational acceleration.

Plugging in the given values, we get 1.44 = 2π√(0.510/g).

Now, let's solve for g. Dividing both sides by 2π and then squaring both sides, we get (1.44/2π)^2 = 0.510/g.

Now we have g = 0.510 / ((1.44/2π)^2).

Calculating this out, we get the experimental value of the gravitational acceleration to be approximately 9.73 m/s².

Alright, next up we have the angular speed of the pendulum. The angular speed, denoted by ω, is simply the 2π divided by the period (ω = 2π/T).

Plugging in the given value for the period, we get ω = 2π/1.44.

Calculating this out, we get the angular speed of the pendulum to be approximately 4.36 rad/s.

So there you have it, my friend. The experimental value of the gravitational acceleration is approximately 9.73 m/s² and the angular speed of the pendulum is approximately 4.36 rad/s. Have fun swinging those equations around!

To determine the experimental value of the gravitational acceleration and the angular speed of the pendulum, we can use the formula for the period of a simple pendulum:

T = 2π√(L / g)

where T is the period, L is the length of the pendulum, and g is the gravitational acceleration.

Given that the length of the pendulum is 0.510 meters and the period is 1.44 seconds, we can rearrange the formula to solve for g:

g = (4π²L) / T²

Substituting the given values, we have:

g = (4π² * 0.510) / (1.44)^2

Calculating this expression will give us the experimental value of the gravitational acceleration.