The equation for the vibration of a string, fixed at both ends vibrating in its third harmonic, is given by ` y = (0.4 cm) sin[(0.314 cm^-1) x] cos[f(600pis^-1)t]`.(a) What is the frequency of vibration ? (b) What are the positions of the nodes ? (c) What is the length of the string ? (d) What is the wavelength and the speed of two travelling waves that can interfere to give this vibration ?

Answer

a) 300Hz
b) 0, 10, 20, 30
c) 30cm
d) 60m/sec

Explanation

the stationary wave equation is givne by
`y=(0.4cm)sin[(0.314cm^-1)x]cos[(6.00pis^-1)t]`
a. `omega=600pi`
`rarr 2pif=600pi`
`gt f=300Hz`
Wavelength
`lamda=(2pi)/0.314=((2xxx3.14))/0.314`
`=20cm`
b. therefoe Nodes are located at 0, 10 cm, 20cm, 30cm.
c. Length of the string
`=(3lamda)/a=(3xx200/2=30cm`
d. `y=0.4sin(0.314x)cos(600pit)`
`=0.4sin{(pi/10)x}cos(600pit)`
Since `lamda` and `v` are the wavelength and velocity of the wave of that interfere to give this vibration.
`lamda=20cm`
`v=omega/k=(600pi)/((pi/10))`
=6000cm/sec=60m/s`