AP Physics 1 Exam Question:

Consider a wave traveling along a stretched string. The wave has a frequency of 500 Hz and a wavelength of 0.2 m. The amplitude of the wave is 0.15 m.

a) Calculate the wave speed.

b) Determine the period of the wave.

c) Find the equation representing the wave.

d) If the wave is heading towards the +x direction, calculate the angular frequency of the wave.

e) Suppose the wave encounters a region where the string is less tense and the tension is reduced to 50% of its initial value. In this new region, calculate the wave speed and the wavelength.

Solution:

a) The wave speed (v) can be calculated using the formula:

v = f * λ

where f = frequency and λ = wavelength.

Therefore, substituting the given values:

v = 500 Hz * 0.2 m = 100 m/s

The wave speed is 100 m/s.

b) The period (T) of a wave is the reciprocal of its frequency:

T = 1 / f

Substituting the given frequency:

T = 1 / (500 Hz) = 0.002 s

The period of the wave is 0.002 s.

c) The equation representing a wave traveling along a stretched string can be written as:

y(x, t) = A * sin(kx - ωt)

where A = amplitude, k = wave number, x = displacement, t = time, and ω = angular frequency.

Substituting the given values:

y(x, t) = 0.15 * sin(2π/0.2 * x - ωt)

d) The angular frequency (ω) of a wave can be calculated using the formula:

ω = 2πf

Substituting the given frequency:

ω = 2π * 500 Hz = 1000π rad/s

The angular frequency is 1000π rad/s.

e) In the region where the string tension is reduced to 50% of its initial value:

The wave speed (v') can be calculated using the formula:

v' = √(T' / μ')

where T' = tension and μ' = mass per unit length.

Since the tension is reduced to 50% of its initial value, T' is equal to 0.5T.

The mass per unit length (μ) of the string remains constant.

Therefore,

v' = √((0.5T) / μ) = 0.707v = 0.707 * 100 m/s ≈ 70.7 m/s

The wave speed in the new region is approximately 70.7 m/s.

The wavelength (λ') in the new region can be calculated using the formula:

λ' = v' / f

Substituting the values:

λ' = (70.7 m/s) / (500 Hz) ≈ 0.141 m

The wavelength in the new region is approximately 0.141 m.