Answer:
a
The speed of wave is
![v_1 = 129.1 \ m/s](https://img.qammunity.org/2021/formulas/physics/college/d67ta0ciqhiqd7wpxmzmozte889oky4mb9.png)
b
The new speed of the two waves is
![v = 129.1 \ m/s](https://img.qammunity.org/2021/formulas/physics/college/dzbgaos99tbnqn2093jwltkyt953550ufg.png)
Step-by-step explanation:
From the question we are told that
The mass of the string is
![m = 60 \ g = 60 *10^(-3) \ kg](https://img.qammunity.org/2021/formulas/physics/college/sfuwp423nq35ljccpozs03anomfl47eruk.png)
The length is
![l = 2.0 \ m](https://img.qammunity.org/2021/formulas/physics/college/300ykqwoel9edqvuvfky2m3q4epgvfa17k.png)
The tension is
![T = 500 \ N](https://img.qammunity.org/2021/formulas/physics/college/pacxur2p292mz8ml1wv3i3feap5sr70cew.png)
Now the velocity of the first wave is mathematically represented as
![v_1 = \sqrt{ (T)/(\mu) }](https://img.qammunity.org/2021/formulas/physics/college/xeambd8xhoxfa2th0b8w45q5blnkbljrk5.png)
Where
is the linear density which is mathematically represented as
![\mu = (m)/(l)](https://img.qammunity.org/2021/formulas/physics/college/50qs4gm1lzf7civrnms7zzco4eyz53k3xu.png)
substituting values
![\mu = ( 60 *10^(-3))/(2.0 )](https://img.qammunity.org/2021/formulas/physics/college/xhhhrlrs29wzexumme869clquxnk11v0qu.png)
![\mu = 0.03\ kg/m](https://img.qammunity.org/2021/formulas/physics/college/lk3tlvbfsaquexxigqzdihkdt01sedbk5b.png)
So
![v_1 = \sqrt{ (500)/(0.03) }](https://img.qammunity.org/2021/formulas/physics/college/v2wnrdfqcjuaidttm2hewh4f915eg79o3g.png)
![v_1 = 129.1 \ m/s](https://img.qammunity.org/2021/formulas/physics/college/d67ta0ciqhiqd7wpxmzmozte889oky4mb9.png)
Now given that the Tension, mass and length are constant the velocity of the second wave will same as that of first wave (reference PHYS 1100 )