Solution
Given,
Length of string (l) = 1.5 m
Density of steel (ρ) = 7800 kg/m^3
Young's modulus (Y) = 2 x 10^11 N/m^2
Strain = 1% = 0.01
Frequency of second mode of vibration (f1) = ?
Now we have,
\(Y=\frac{stress}{strain}\)
Or, \(stress=Y\times strain\)
Or, \(\frac{Tension}{Area}=2\times10^{11}\times\frac1{100}\)
Or, \(\frac TA=2\times10^9\) ------- (i)
Then, for fundamental mode, we have
\(f=\frac1{2l}\sqrt{\frac T{M/L}}=\frac1{2l}\sqrt{\frac{LT}M}=\frac1{2l}\sqrt{\frac{LT}{A\times L\times\rho}}=\frac1{2l}\sqrt{\frac T{A\times\rho}}\)
\(=\frac1{2\times1.5}\sqrt{\frac{2\times10^9}{7800}}=168.79\) Hz
Hence, the required frequency is 168.79 Hz
For the second mode of vibration
\(f_2=2f=2\times168.79=337.6\) Hz