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 0.383 m

. The maximum transverse acceleration of a point at the middle of the segment is 8100 m/s2
and the maximum transverse velocity is 3.20 m/s
. A) What is the amplitude of this standing wave? B) What is the wave speed for the transverse traveling waves on this string?

Answer 1

The amplitude of the standing wave is 0.0533 m and the wave speed for the transverse traveling waves on the string is 338.679 m/s.

Step-by-step explanation:

To find the amplitude of the standing wave, we can use the formula for maximum transverse velocity. The maximum transverse velocity is equal to the amplitude multiplied by the wave speed. Rearranging the formula, we can solve for the amplitude and substitute the given values:

Amplitude = Maximum Transverse Velocity / Wave Speed
Amplitude = 3.20 m/s / 60.00 m/s

= 0.0533 m

To find the wave speed for the transverse traveling waves on the string, we can use the formula: wave speed = frequency × wavelength. Since it is the fundamental mode, the length of the segment of the string that is free to vibrate is equal to half a wavelength:

Wave Speed = 2 × Length / Period
Wave Speed = 2 × 0.383 m / (1/440 Hz)

= 338.679 m/s