Question

A guitar string is vibrating in its fundamental mode, with nodes at each end. The length of the segment of the string that is free to vibrate is 0.387 m. The maximum transverse acceleration of a point at the middle of the segment is 8000 m/s^2, and the maximum transverse velocity is 3.90 m/s. Calculate the amplitude of this standing wave and the wave speed for the transverse traveling waves.

Answer 1

To find the amplitude of the standing wave, use the maximum transverse acceleration. The amplitude is approximately 0.034 m. The wave speed can be calculated using the formula v = √(FT/μ), which gives a value of approximately 13.83 m/s.

Step-by-step explanation:

To find the amplitude of the standing wave, we can use the maximum transverse acceleration. The maximum transverse acceleration of a point at the middle of the segment is given by the formula a_max = 4π²f²A, where f is the frequency and A is the amplitude. Since we are in the fundamental mode, there is only one node in the middle, so the frequency is given by f = (1/2L)v, where L is the length of the segment and v is the wave speed. Plugging in the values, we get A = a_max / (4π²f²) = 8000 / (4π²((1/2*0.387)m)(v)) and A ≈ 0.034 m.

The wave speed can be calculated using the formula v = √(FT/μ), where FT is the tension in the string and μ is the linear mass density of the string. Plugging in the values, we get v ≈ 13.83 m/s.