Your boss at the Cut-Rate Cuckoo Clock Company wants to save money on materials. You are asked to reduce the balance wheel, which is cylindrical, dimensions to one-third their original values. Assume the new wheel has the same density and the same coil spring (thus the same torsion constant) as the original.

Would the frequency of the angular simple harmonic motion (SHM) of the balance wheel increase or decrease if the dimensions of the balance wheel were changed as described?

By what factor would the frequency of the angular SHM of the balance wheel change?

a) Angular velocity decreases

b) for L / 3. the frequency changes by 1 / √3

for R/√3 the frequency changes by √ 9

Explanation:

As in the problem they indicate that we have a torsion constant, we have a torsion pendulum, with angular velocity

w = √ I / k

The angular momentum of a cylinder with an axis of rotation passing through its center is

I = ½ m R²

We replace

w = √ ½ m R² / k

Angular velocity and frequency are related.

w = 2π f

f = w/2π

Let's analyze what happens with the requested changes. We will use the subscript zero for the initial dimensions. The dimensions have been reduced to 1/3 of the original values, but maintaining the wheel density

ρ = m₀ / V₀

With the new values

ρ = (m₀ / 3) / (V₀ / 3)

So that the density does not change, let's look for what change we must make in the volume of the cylinder

V = π r² L

We can make two changes

1) L = L₀/3, maintaining the same radius

The angular velocity is left with these changes

w' = √ ½ m₀/3 R₀² / k = 1 / √3 √ ½ m₀ R₀ / k

w' = w₀ / √3

Therefore the angular velocity decreases

Frequency is

f' / 2pi = 1 / √3 f₀ / 2pi

f' = 1 / √3 f₀

2) R = √3 R₀, maintaining the same length

Angular velocity

w' = √ ½ m₀ / 3 (R₀² / 3) / k

w = 1 / √9 √ ½ m₀R₀² / k

w ' = 1 / √9 w₀

Angular velocity decreases

The angular frequency is

f ' = f₀ / √ 9

The frequency also decreases