Calculate the period of oscillation of a simple pendulum of length 1.8 m, with a bob of mass 2.2 kg. What assumption is made in this calculation?
(g = 9.8 m s - 2)
If the bob of this pendulum is pulled aside a horizontal distance of 20cm and
released, what will be the values of (i) the kinetic energy and (ii) the velocity of the bob at the lowest point of the swing?
student
To calculate the period of oscillation of a simple pendulum, we can use 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.
Given:
Length of the pendulum (L) = 1.8 m
Acceleration due to gravity (g) = 9.8 m/s^2
Substituting these values into the formula:
T = 2π√(1.8/9.8)
T ≈ 2π√0.1837
T ≈ 2π * 0.4286
T ≈ 2.69 seconds
The assumption made in this calculation is that the amplitude of oscillation is small. This assumption is known as the small-angle approximation.
Now, let's calculate the values of (i) the kinetic energy and (ii) the velocity of the bob at the lowest point of the swing.
(i) The kinetic energy (KE) can be calculated using the formula:
KE = (1/2)mv^2
where m is the mass of the bob and v is its velocity.
Given:
Mass of the bob (m) = 2.2 kg
Since the bob is at its lowest point of swing, all of its potential energy is converted into kinetic energy. Therefore, the kinetic energy is equal to the total energy of the system at this point.
(ii) The velocity (v) can be calculated using the conservation of mechanical energy:
Potential energy (PE) + Kinetic energy (KE) = Total energy (E)
Since the bob is at its lowest point, its potential energy is zero. Therefore:
KE = E
Substituting the values:
KE = (1/2)(2.2 kg)(v^2)
At the lowest point, all the potential energy is converted into kinetic energy, so:
KE = mgh, where h is the height from the lowest point of the swing, which is 20 cm or 0.20 m.
So, KE = (2.2 kg)(9.8 m/s^2)(0.20 m)
Now, solving for KE:
KE = 4.31 Joules
Therefore:
(i) The kinetic energy of the bob at the lowest point of the swing is approximately 4.31 Joules.
(ii) The velocity of the bob at the lowest point of the swing can be found using the equation:
KE = (1/2)mv^2
Substituting the values:
4.31 J = (1/2)(2.2 kg)(v^2)
Now, let's solve for v:
v^2 = (2 * 4.31 J) / (2.2 kg)
v^2 ≈ 3.92 m^2/s^2
v ≈ √(3.92)
v ≈ 1.98 m/s
Therefore, (ii) The velocity of the bob at the lowest point of the swing is approximately 1.98 m/s.
To calculate the period of oscillation of a simple pendulum, you can use 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.
In this case, the length of the pendulum (L) is given as 1.8 m and g is given as 9.8 m/s^2. Plugging these values into the formula:
T = 2π√(1.8/9.8)
Simplifying:
T ≈ 2π√0.183
T ≈ 2π * 0.428
T ≈ 2.69 s (rounded to two decimal places)
Now, let's address the assumption made in this calculation. The formula assumes that the pendulum undergoes small oscillations, meaning that the angle of displacement is small. If the angle becomes too large, the pendulum will not follow simple harmonic motion and the formula will no longer be accurate.
Moving on to the second part of the question, let's calculate the values of kinetic energy and velocity of the bob at the lowest point of the swing after being released from a horizontal distance of 20 cm.
To calculate the kinetic energy of the bob, we'll need to know its velocity.
The potential energy at the highest point is converted into kinetic energy at the lowest point. Assuming no energy losses due to friction or air resistance:
Potential Energy (PE) = Kinetic Energy (KE)
PE = mgh
where m is the mass of the bob, g is the acceleration due to gravity, and h is the height of the pendulum (in this case, the horizontal distance of 20 cm, which is equivalent to 0.2 m).
KE = PE = mgh
KE = (2.2 kg)(9.8 m/s^2)(0.2 m)
KE ≈ 4.31 J (rounded to two decimal places)
The velocity of the bob at the lowest point can be calculated using the equation:
KE = (1/2)mv^2
where KE is the kinetic energy, m is the mass of the bob, and v is the velocity.
Rearranging the equation, we get:
v = √(2KE/m)
Substituting the values:
v = √(2 * 4.31 J / 2.2 kg)
v ≈ 1.70 m/s (rounded to two decimal places)
So, the values are (i) the kinetic energy is approximately 4.31 J, and (ii) the velocity of the bob at the lowest point is approximately 1.70 m/s.