Answer:
Maximum speed for a point on the string at anti node will be 22.6 m/sec
Explanation:
We have given length of string L = 3.5 m
For 7th harmonic length of the string [tex]L=\frac{7\lambda }{2}[/tex]
So [tex]\lambda =\frac{2L}{7}[/tex]
Speed of the wave in the string is 150 m/sec
Frequency corresponding to this wavelength [tex]f=\frac{v}{\lambda }=\frac{7v}{2L}[/tex]
So angular frequency will be equal to [tex]\omega =2\pi f=2\pi \times \frac{7v}{2L}=2\times 3.14\times \frac{7\times 150}{2\times 3.5}=942rad/sec[/tex]
Maximum speed is equal to [tex]v_m=A\omega =0.024\times 942=22.60m/sec[/tex]
So maximum speed for a point on the string at anti node will be 22.6 m/sec
