A horizontal plank (m = 2.0 kg, L = 1.0 m) is pivoted at one end. A spring. (k = 1.0 x 10. 3. N/m) is attached at the other end find the angular frequence for small oscillations?
There is nothing wrong with this answer to what I know but here is how my comrade and I did it:
The angular frequency of a simple harmonic oscillator is as follows:
w=sqrt((k/I))
The system is called a torsional pendulum. and if one replaces I, the moment of inertia, with 1/3(m)(r^2) you will get the sqrt(k/(1/3(m)(r^2). and that leads to sqrt(3k/m(r^2)) and well since 1 =r^2 its sqrt(3k/m). Enjoy
There is a dynamic relationship between torque about the pivot and the rate of change of angular momentum, that can be written as a differential equation
If the linear deflection at the attachment point is X, the angular deflection A is
A = X/L
The angular momentum is
I dA/dt wherre I is the moment of inertia, (1/3) m L^2. The equation of motion is
Torque = -k X*L = -k L^2 A =
d/dt I dA/dt = (1/3) mL^2 d^2A/dt^2
d^2A/dt^2 + 3 k/m A = 0
The angular oscillation frequency that results from this differential equation is
w = sqrt (3k/m)
(The length cancels out). Note how this resembles the spring-mass natural oscillation frequency withough the hinge, which would be sqrt(k/m)
To find the angular frequency for small oscillations of the system, we can use the equation:
ω = √(k / I)
where:
- ω represents the angular frequency
- k is the spring constant
- I is the moment of inertia of the plank
The moment of inertia of a thin rod rotating about its end can be given by the formula:
I = (1/3) * m * L^2
where:
- m represents the mass of the plank
- L is the length of the plank
In this case, the mass of the plank (m) is given as 2.0 kg, and the length of the plank (L) is given as 1.0 m. The spring constant (k) is given as 1.0 x 10^3 N/m.
Plugging the values into the formula for moment of inertia, we can calculate I:
I = (1/3) * 2.0 kg * (1.0 m)^2
I = (1/3) * 2.0 kg * 1.0 m^2
I = (2/3) kg * m^2
Now, we can substitute the values of k and I into the equation for angular frequency:
ω = √(1.0 x 10^3 N/m / (2/3) kg * m^2)
Simplifying further:
ω = √((1.0 x 10^3 N/m) * (3/2 kg * m^2))
ω = √(1500 (kg * m^2 / N))
ω = √(1500 s^-2)
Finally, calculating the square root:
ω ≈ 38.73 s^-1
Therefore, the angular frequency for small oscillations in this system is approximately 38.73 s^-1.