Answer:
[tex]a. \ f_1=7.9057Hz\\\\b. \ f_2=15.8114Hz\\\\c. \ f_3=23.7171Hz[/tex]
Explanation:
a. The wire's length is 10m long and has a mass 100g and a tension of 250N.
Frequency is given by the equation:
[tex]f=\frac{nv}{2L}\\\\=\frac{n}{2L}\sqrt{\frac{t}{\mu}[/tex] #where t=250N*10=2500N, [tex]\mu=0.1kg[/tex]
#substitute for actual values for the lowest frequency.
[tex]F=\frac{n}{2\times 10m}\times \sqrt{2500N/0.1kg}\\=\frac{1}{2\times 10m}\times \sqrt{2500N/0.1kg}\\=7.9057Hz[/tex] #n=1, lowest frequency
Hence, the lowest frequency for standing waves is 7.9057Hz
b.The wire's length is 10m long and has a mass 100g and a tension of 250N.
Frequency is given by the equation:
[tex]f=\frac{nv}{2L}\\\\=\frac{n}{2L}\sqrt{\frac{t}{\mu}[/tex] #where t=250N*10=2500N,[tex]\mu=0.1kg[/tex]
#The second lowest frequency happens at [tex]n=2[/tex]:
[tex]F=\frac{n}{2\times 10m}\times \sqrt{2500N/0.1kg}\\=\frac{2}{2\times 10m}\times \sqrt{2500N/0.1kg}\\=15.8114Hz[/tex]
Hence, the second lowest frequency is 15.8114Hz
c.Given that the wire's length is 10m long and has a mass 100g and a tension of 250N.
Frequency is given by the equation:
[tex]f=\frac{nv}{2L}\\\\=\frac{n}{2L}\sqrt{\frac{t}{\mu}[/tex] #where t=250N*10=2500N,[tex]\mu=0.1kg[/tex]
The third lowest frequency happens at [tex]n=3[/tex]
[tex]F=\frac{n}{2\times 10m}\times \sqrt{2500N/0.1kg}\\=\frac{3}{2\times 10m}\times \sqrt{2500N/0.1kg}\\=23.7171Hz[/tex]
Hence, the third lowest frequency is 23.7171Hz
