Step-by-step explanation:
The wave equation which is propagating down a string of unknown material is given by :
................(1)
The general equation of wave is given by :
..............(2)
Comparing equation (1) and (2) we get:
(a)
![\omega=6](https://img.qammunity.org/2020/formulas/physics/college/60qgqlqu1r5sdi0x1sjt0hdryscewtmbct.png)
![2\pi f=6](https://img.qammunity.org/2020/formulas/physics/college/cb82z9440pl7a5fiua5de5ryhalwq2o7ag.png)
f = 0.954 Hz
(b) Radian frequency,
![\omega=6\ rad/s](https://img.qammunity.org/2020/formulas/physics/college/vamcz1xnawv2qsu0rub8g56ne6ijax2128.png)
(c) k = 7
![(2\pi)/(\lambda)=7](https://img.qammunity.org/2020/formulas/physics/college/bhexsioksh408y07wnjly76lkovbxl2zd4.png)
![\lambda=0.89\ m](https://img.qammunity.org/2020/formulas/physics/college/2fsaqwxv87hvzxh0hnsif5l1l48bn5svp2.png)
(d) Wave number,
![k=5\ m^(-1)](https://img.qammunity.org/2020/formulas/physics/college/cs6fcd0b36j363to3anehs6mznwg6n5i0u.png)
(e) Speed of sound,
![v=f* \lambda](https://img.qammunity.org/2020/formulas/physics/college/46mjvyrfm85si9wcfzkms2zzno9mgoxpke.png)
![v=0.954* 0.89](https://img.qammunity.org/2020/formulas/physics/college/zwfhg3wuzcqp58485c572lhubvd40rkx3r.png)
v = 0.849 m/s
Hence, this is the required solution.