Question

Shows two masses hanging from a steel wire. the mass of the wire is 57.5 g. a wave pulse travels along the wire from point 1 to point 2 in 23.5 ms.

Answer 1

The wave speed in section A is 20.00 m/s and the wave speed in section B is 16.00 m/s.

Step-by-step explanation:

To find the wave speed in section A and section B, we need to use the formula: wave speed = sqrt(tension / linear mass density).

For section A, we have: wave speed in section A = sqrt(5.00 N / 0.0025 kg/m) = 20.00 m/s.

For section B, we have: wave speed in section B = sqrt(4.00 N / 0.0025 kg/m) = 16.00 m/s.

Answer 2

The wave speed in section A is 257.34 m/s, and the wave speed in section B is 265.78 m/s.

Step-by-step explanation:

To find the wave speed in section A, we need to use the formula v = √(T/μ). Plugging in the given values, we get v = √(600 N / 0.0085 kg/m) = 257.34 m/s. To find the wave speed in section B, we can use the same formula with the tension in string 2 and the same linear mass density μ. Plugging in the values, we get v = √(700 N / 0.0085 kg/m) = 265.78 m/s.