2 simple oscilliation problems.

[physicszman]

1) You are a new employee at the Cut-Rate Cuckoo Clock company. The boss asks you what would happen to the frequency of the angular simple harmonic motion of the balance wheel if it had the same density and the same coil spring (thus the same torsion constant), but all the balance wheel dimensions were reduced by a factor of 2.56 to save material.

a)You tell the boss (a) it is a dumb idea that will not work and (b) because the period will change by a factor of ?

i use the formula t = 2pie x sqrt(L/M)
then just change it to t = 2pie x sqrt(L/(1/2.56M))

but i have trouble finding the ratio?

b)By what factor must to the torsion constant be changed to make the smaller balance wheel oscillate at the same frequency?

*need formula please

2) A uniform circular disk whose radius R is 13.1 cm is suspended as a physical pendulum from a point on its rim. (The disk pivots about a loose-fitting nail driven through its face very close to the rim.) What is the period of oscillation?

*need formula please!


If you guys coudl just provide me with a clearers visual description and a dormula of what is happening I can do the math. Any help or suggestions are appreciated thank you!

 

[ShawnD]

i use the formula t = 2pie x sqrt(L/M)
then just change it to t = 2pie x sqrt(L/(1/2.56M))

Where did you get this formula from? What is M? Do the units even work out?

I'm not entirely sure what the system looks like, and seeing it makes a huge difference.

2) A uniform circular disk whose radius R is 13.1 cm is suspended as a physical pendulum from a point on its rim. (The disk pivots about a loose-fitting nail driven through its face very close to the rim.) What is the period of oscillation?

*need formula please!

Draw where the center of mass is, draw where the center of rotation is, then go at it from there.

t = 2\pi \sqrt{\frac{L}{g}}

 

[M98Ranger]

a) In response to the first problem, you can tell your boss that it is not a good idea because the period of the simple harmonic motion will change by a factor of 2.56. This is because the period is directly proportional to the square root of the length and inversely proportional to the square root of the mass. Therefore, if the length is reduced by a factor of 2.56, the period will increase by the same factor (since 1/sqrt(2.56) = 1/1.6 = 0.625).

b) To make the smaller balance wheel oscillate at the same frequency, the torsion constant must be increased by a factor of 2.56. This is because the frequency of simple harmonic motion is directly proportional to the square root of the torsion constant.

The formula for the period of a simple harmonic motion is T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. In this case, the length of the balance wheel is reduced by a factor of 2.56, so the new period will be T' = 2π√(L/2.56g). To make T' equal to T, we need to increase the torsion constant by a factor of 2.56.

2) For the second problem, the formula for the period of a physical pendulum is T = 2π√(I/mgd), where I is the moment of inertia of the pendulum, m is the mass, g is the acceleration due to gravity, and d is the distance from the pivot point to the center of mass. In this case, the disk is a uniform circular disk, so the moment of inertia is I = 1/2mr^2, where r is the radius of the disk. Plugging in the given values, we get T = 2π√(1/2mr^2/mgd) = 2π√(r^2/gd).

Since the radius of the disk is given as 13.1 cm, we can convert it to meters by dividing by 100, giving us r = 0.131 m. The distance from the pivot point to the center of mass is equal to the radius of the disk, so d = 0.131 m. Plugging these values into the formula, we get