Given,
The initial length of the string, L₁=0.50 m
The tension on the string, T₁=2.0×10² N
The initial fundamental frequency of the string, f₁=400 Hz
The length of the string after it was shortened, L₂=0.35 m
The increased tension on the string, T₂=4.0×10² N
The fundamental frequency of the string before it was shortened is given by,
[tex]f_1=\frac{\sqrt[]{\frac{T_1}{\mu}}}{2L_1}[/tex]Where μ is the mass per unit length of the string.
On rearranging the above equation,
[tex]\begin{gathered} 4L^2_1f^2_1=\frac{T_1}{\mu} \\ \Rightarrow\mu=\frac{T_1}{4L^2_1f^2_1} \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} \mu=\frac{2.0\times10^2}{4\times0.50^2\times400^2} \\ =1.25\times10^{-3}\text{ kg/m} \end{gathered}[/tex]The fundamental frequency after the string was shortened is given by,
[tex]f_2=\frac{\sqrt[]{\frac{T_2}{\mu}}}{2L_2}[/tex]On substituting the known values,
[tex]\begin{gathered} f_2=\frac{\sqrt[]{\frac{4\times10^2}{1.25\times10^{-3}}}}{2\times0.35} \\ =808.1\text{ Hz} \end{gathered}[/tex]Thus the fundamental frequency after the string is shortened and the tension is increased is 808.1 Hz
Hello! So I need some help with this question. I don’t quite understand it and I did have another tutor figure it out and the answer was wrong. It’s not his answer of 58.04. Can you help me?
ANSWER:
58 m/s
STEP-BY-STEP EXPLANATION:
Given:
Initial velocity (u) = 13 m/s
The correct answer is 58 m/sDistance (d) = 400 m
Acceleration (a) = 4 m/s²
We apply the following formula to determine the final velocity:
[tex]\begin{gathered} v^2=u^2+2as \\ \\ \text{ We replacing:} \\ \\ v^2=13^2+2(4)(400) \\ \\ v^2=169+3200 \\ \\ v=\sqrt{3369} \\ \\ v=58.04\cong58\text{ m/s} \end{gathered}[/tex]The correct answer is 58 m/s