Vibration with frequency 6.00x10^2 Hz are established on a 1.33m length of string that is clamped at both ends. The speed wave on the string is 4.00x10^2 m/s. How far from either end of the string does the first node occur?

Answer 1

To find the distance from either end of the string where the first node occurs, we can use the formula for the wavelength of a wave on a string:

wavelength = (2L) / n

where:
- wavelength is the distance between two consecutive nodes or antinodes
- L is the length of the string
- n is an integer representing the number of nodes or antinodes

We are given:
- frequency = 6.00x10^2 Hz
- length of the string (L) = 1.33 m
- wave speed = 4.00x10^2 m/s

The wavelength can be calculated using the formula:
wavelength = wave speed / frequency

wavelength = (4.00x10^2 m/s) / (6.00x10^2 Hz)
wavelength = 0.67 m

The first node represents a half-wavelength, so it occurs at one-fourth of a full wavelength from either end of the string.

distance from either end = (wavelength / 4)
distance from either end = (0.67 m) / 4
distance from either end = 0.17 m

Therefore, the first node occurs at a distance of 0.17 m from either end of the string.