an ethernet cable is 4 m long and has a mass of 0.2 kg . a transverse wave is produced by plucking one end of the taut cable. the pulse makes 4 trips down and back along the cable in 0.8 sec , what is the tension in the cable?

Question

anonymous · Answer

Let "$T$" be the tension in the cable, and then the vertical component of the cable tension is: 

                                               $F_v = T×Sin(30) $

Now, try to solve it using the formula.. @YamadaTasnim

anonymous · Answer

$$trips / time= (1/2L) \sqrt{T/(kg/m)} $$
i use this formula.. got it right! @Persia

michele_laino · Answer

the emitted frequency $ \large \nu $ by the cable is given by the subsequent formula:
$$\Large \nu  = \frac{1}{{2L}}\sqrt {\frac{T}{\rho }} \;\left( {{\text{Hz}}} \right)$$
where $ \large \rho $ is the linear density of the cable, and $ \large T $ is the tension which is acting on the cable