## College Physics (4th Edition)

We can write an expression for the wave speed along a string: $v = \sqrt{\frac{F}{m/L}} = \sqrt{\frac{F~L}{m}}$ We can find an expression for the tension: $\frac{v}{\lambda} = f$ $\frac{\sqrt{\frac{F~L}{m}}}{2L} = f$ $\sqrt{\frac{F~L}{m}} = 2L~f$ $\frac{F~L}{m} = (2L~f)^2$ $F = 4~m~L~f^2$ We can find the required tension $F'$ to decrease the fundamental frequency by 4.0%: $F' = 4~m~L~(0.96~f)^2$ $F' = 0.96^2 \times 4~m~L~f^2$ $F' = 0.96^2 \times F$ $F' = 0.922 \times F$ The tension should be reduced by 7.8%