Answer:
5 m/s
Explanation:
The speed of a wave in a string is related to the tension in the string by the equation
[tex]v=\sqrt{\frac{T}{\mu}}[/tex]
where
v is the speed of the wave
T is the tension in the string
[tex]\mu[/tex] is the linear density of the string
We can rewrite the equation as
[tex]\frac{\sqrt{T}}{v}=\sqrt{\mu}[/tex]
In this problem, the tension in the string is changed; however, its linear mass density remains constant. So we can write:
[tex]\frac{\sqrt{T_1}}{v_1}=\frac{\sqrt{T_2}}{v_2}[/tex]
where:
T1 = 40 N is the initial tension in the string
v1 = 10 m/s is the initial speed of the wave
T2 = 10 N is the final tension in the string
Solving for v2, we find the final speed of the wave:
[tex]v_2=v_1 \sqrt{\frac{T_2}{T_1}}=(10)\sqrt{\frac{10}{40}}=5 m/s[/tex]
