Complete Question
The speed of a transverse wave on a string of length L and mass m under T is given by the formula
![v=\sqrt{(T)/((m/l))}](https://img.qammunity.org/2022/formulas/physics/college/okqbsn4ccuvbbgjqiu3uu82ld0qug5fqvd.png)
If the maximum tension in the simulation is 10.0 N, what is the linear mass density (m/L) of the string
Answer:
![(m/l)=(10)/(V^2)](https://img.qammunity.org/2022/formulas/physics/college/5ptib1p5cskwvh0ibt4vz76tnfj0qbszaq.png)
Step-by-step explanation:
From the question we are told that
Speed of a transverse wave given by
![v=\sqrt{(T)/((m/l))}](https://img.qammunity.org/2022/formulas/physics/college/okqbsn4ccuvbbgjqiu3uu82ld0qug5fqvd.png)
Maximum Tension is
![T=10.0N](https://img.qammunity.org/2022/formulas/physics/college/jqkmtlfcwaby2bglvpvx1eck2adrmv7jdk.png)
Generally making
subject from the equation mathematically we have
![v=\sqrt{(T)/((m/l))}](https://img.qammunity.org/2022/formulas/physics/college/okqbsn4ccuvbbgjqiu3uu82ld0qug5fqvd.png)
![v^2=(T)/((m/l))](https://img.qammunity.org/2022/formulas/physics/college/jdb4a9qub63rdhc2st7r7vpnfq4qu9okwg.png)
![(m/l)=(T)/(V^2)](https://img.qammunity.org/2022/formulas/physics/college/lcd5jko1ilnk9k9ojih1jcajoie6w1p9ne.png)
![(m/l)=(10)/(V^2)](https://img.qammunity.org/2022/formulas/physics/college/5ptib1p5cskwvh0ibt4vz76tnfj0qbszaq.png)
Therefore the Linear mass in terms of Velocity is given by
![(m/l)=(10)/(V^2)](https://img.qammunity.org/2022/formulas/physics/college/5ptib1p5cskwvh0ibt4vz76tnfj0qbszaq.png)