- #1
jimisreincarn
- 3
- 0
1. If the length of the tube used in our experiment is 4 m, how many resonances would you observe when a tuning fork of frequency 256HZ is used?
2. v=f[tex]\lambda[/tex] ; L = (1/4)(2n+1)[tex]\lambda[/tex]
3. 340m/s = 256Hz[tex]\lambda[/tex] [tex]\lambda[/tex] = 1.328125m
4 = (1/4)(2n+1)(1.328125m)
12.05 = 2n+1
5.5 = n
2. v=f[tex]\lambda[/tex] ; L = (1/4)(2n+1)[tex]\lambda[/tex]
3. 340m/s = 256Hz[tex]\lambda[/tex] [tex]\lambda[/tex] = 1.328125m
4 = (1/4)(2n+1)(1.328125m)
12.05 = 2n+1
5.5 = n