AP Physics 2 Exam question:

A wave propagates along a string with a speed of 20 m/s. The string has a tension of 50 N and a mass per unit length of 0.02 kg/m.

a) Calculate the wavelength and frequency of the wave if the string has a linear density of 0.01 kg/m².

b) The wave travels from a medium with a refractive index of 1.5 to a medium with a refractive index of 1.2. Determine the angle of incidence for total internal reflection to occur.

Answer:

a) The relationship between wave speed, wavelength, and frequency can be expressed as:

v = λf

Where:

• v is the wave speed (20 m/s in this case)
• λ is the wavelength (to be determined)
• f is the frequency (to be determined)

Since the wave propagates along a string, we can make use of the wave equation specific to waves on a string:

v = √(T/μ)

Where:

• T is the tension in the string (50 N in this case)
• μ is the linear mass density (0.02 kg/m in this case)

We can rearrange the equation above to solve for μ as:

μ = T / v²

Plugging in the given values, we have:

μ = (50 N) / (20 m/s)² = 0.125 kg/m

The linear mass density is also related to the linear density, µ, as:

μ = µ / λ

We can rearrange this equation to solve for the wavelength, λ, as:

λ = µ / μ

Plugging in the given values, we have:

λ = (0.01 kg/m²) / (0.125 kg/m) = 0.08 m (or 8 cm)

Finally, we can find the frequency, f, by rearranging the equation mentioned earlier:

f = v / λ

Plugging in the given values, we have:

f = (20 m/s) / (0.08 m) = 250 Hz

Therefore, the wavelength of the wave is 0.08 m (or 8 cm) and the frequency is 250 Hz.

b) The critical angle for total internal reflection can be determined using Snell's Law:

n₁sinθ₁ = n₂sinθ₂

Where:

• n₁ is the refractive index of the first medium (1.5 in this case)
• θ₁ is the angle of incidence (to be determined)
• n₂ is the refractive index of the second medium (1.2 in this case)
• θ₂ is the angle of refraction (90 degrees for total internal reflection)

Rearranging the equation, we have:

sinθ₁ = (n₂ / n₁)sinθ₂

For total internal reflection to occur, the angle of incidence, θ₁, must be the critical angle, which means the angle of refraction, θ₂, is 90 degrees. Thus:

sinθ₂ = sin 90 = 1

Now we can solve for the angle of incidence, θ₁:

sinθ₁ = (n₂ / n₁)sinθ₂ = (1.2 / 1) * 1 = 1.2

From this equation, we can conclude that since the sine of the angle of incidence cannot be greater than 1, total internal reflection will not occur. Therefore, there is no angle of incidence for total internal reflection in this scenario.