One end of a light spring with force constant 300 N/m is attached to a vertical wall. A light string is tied to the other end of the horizontal spring. The string changes from horizontal to vertical as it passes over a solid pulley of diameter 6.00 cm. The pulley is free to turn on a fixed smooth axle. The vertical section of the string supports a 400 g object. The string does not slip at its contact with the pulley.
(b) Find the frequency of oscillation if the mass of the pulley is 450 g.
(c) Find the frequency of oscillation if the mass of the pulley is 750 g.
SOLVE
YES
PHXSICS
To find the frequency of oscillation, we first need to calculate the effective mass of the system for each case.
(b) Mass of the pulley is 450 g:
The mass of the pulley affects the effective mass of the system since it is part of the oscillating system. We need to consider the moment of inertia of the pulley to calculate its effective mass.
The moment of inertia of a solid disc rotating about an axis passing through its center and perpendicular to its plane is given by the formula:
I = (1/2) * m * r^2,
where I is the moment of inertia, m is the mass of the object, and r is the radius of the pulley.
In this case, the mass of the pulley is 450 g, and the radius is half of the diameter (6.00 cm), so r = 3.00 cm = 0.03 m.
Using the formula, I = (1/2) * 0.450 kg * (0.03 m)^2 = 0.0027 kg m^2.
The effective mass of the system is given by the sum of the masses attached to the string and the effective mass of the pulley:
m_eff = m_object + I * (1 / r^2),
where m_eff is the effective mass, m_object is the mass of the object, and I is the moment of inertia of the pulley.
In this case, the mass of the object is 400 g, so m_object = 0.400 kg.
m_eff = 0.400 kg + 0.0027 kg m^2 * (1 / (0.03 m)^2) = 0.400 kg + 3 kg = 3.400 kg.
The frequency of oscillation is given by the formula:
f = (1 / (2 * pi)) * sqrt(k / m_eff),
where f is the frequency, k is the force constant of the spring, and m_eff is the effective mass of the system.
In this case, the force constant of the spring is 300 N/m.
f = (1 / (2 * pi)) * sqrt(300 N/m / 3.400 kg) = 4.03 Hz (rounded to two decimal places).
Therefore, the frequency of oscillation is approximately 4.03 Hz with a pulley mass of 450 g.
(c) Mass of the pulley is 750 g:
We follow the same steps to calculate the effective mass of the system.
The mass of the pulley is 750 g, so m_pulley = 0.750 kg.
Using the same radius of 0.03 m, the moment of inertia of the pulley is:
I = (1/2) * 0.750 kg * (0.03 m)^2 = 0.003375 kg m^2.
The effective mass of the system is given by:
m_eff = m_object + I * (1 / r^2) = 0.400 kg + 0.003375 kg m^2 * (1 / (0.03 m)^2) = 0.400 kg + 3.75 kg = 4.15 kg.
Using the formula for frequency:
f = (1 / (2 * pi)) * sqrt(k / m_eff) = (1 / (2 * pi)) * sqrt(300 N/m / 4.15 kg) = 3.78 Hz (rounded to two decimal places).
Therefore, the frequency of oscillation is approximately 3.78 Hz with a pulley mass of 750 g.