Learning object 6
- 1. Learning Object #6 A string 0.300m long, with a tension force of 80.0N and a linear mass density of 2.00kg/m. A) Calculate the wave speed in the string. Solution: v=sqrt(tension/linear density) v=6.32m/s B) What is the wavelength of the string if n=6? Solution: f=(n/2L)v Since v is constant, because tension and linear mass doesn’t change, frequency at n=6 is 6 times the frequency at fundamental harmonic. f1=（1/(2*0.30m))v f1=10.5Hz f6=6(10.5Hz)=63.0Hz v=(frequency)(wavelength) (6.32m/s)/(63.0)=0.100m C) If the fundamental frequency increase by 10Hz, with the same string. What factor has to change in order for an increase in fundamental frequency? Solution: Because its the same string, the linear density of the string and the wavelength at the fundamental harmonic shouldn’t change. If the frequency increase, then v=(frequency)(wavelength), so wave speed changes. v=sqrt(tension/linear density) Thus, the tension should increase in order for the frequency at fundamental harmonic to change. D) By how much should the tension increase in order for the frequency at fundamental harmonic to increase by 10Hz? Solution: f1=10.5Hz+10Hz=20.5Hz f1=(1/2L)sqrt(T/u) T=(2L*f1)^2*u T=303N T=303N-80.0N=223N The tension should increase by 223N.