Answer
given,
Length of the string, L = 2 m
speed of the wave , v = 50 m/s
string is stretched between two string
For the waves the nodes must be between the strings
the wavelength is given by
![\lambda = (2L)/(n)](https://img.qammunity.org/2021/formulas/physics/high-school/s1818mhvi0153k4kha66rq7fojs1s276pf.png)
where n is the number of antinodes; n = 1,2,3,...
the frequency expression is given by
![f = n(v)/(2L)](https://img.qammunity.org/2021/formulas/physics/high-school/5kpptgsvjpb0hm9hsptms0cqkyco0i2qp2.png)
now, wavelength calculation
n = 1
![\lambda_1 = (2* 2)/(1)](https://img.qammunity.org/2021/formulas/physics/high-school/yt1a1zpxgfd9etdrgvxn6u9rbl7unyafgu.png)
λ₁ = 4 m
n = 2
![\lambda_2 = (2* 2)/(2)](https://img.qammunity.org/2021/formulas/physics/high-school/kwabpa9bhmd5mnhiark7yvk1hs10ctviup.png)
λ₂ = 2 m
n =3
![\lambda_3 = (2* 2)/(3)](https://img.qammunity.org/2021/formulas/physics/high-school/zb4u6mpa7g7r5k0p9tcetoyb95cdfmvwl7.png)
λ₃ = 1.333 m
now, frequency calculation
n = 1
![f = n(v)/(2L)](https://img.qammunity.org/2021/formulas/physics/high-school/5kpptgsvjpb0hm9hsptms0cqkyco0i2qp2.png)
![f_1 =1* (50)/(2* 2)](https://img.qammunity.org/2021/formulas/physics/high-school/9pkapk6d7spcjpbuzn3gu34zlrbhxivj16.png)
f₁ = 12.5 Hz
n = 2
![f = n(v)/(2L)](https://img.qammunity.org/2021/formulas/physics/high-school/5kpptgsvjpb0hm9hsptms0cqkyco0i2qp2.png)
![f_2 =2* (50)/(2* 2)](https://img.qammunity.org/2021/formulas/physics/high-school/2p067n2j2zmbq2de89g359rypmnwkfvp4c.png)
f₂= 25 Hz
n = 3
![f = n(v)/(2L)](https://img.qammunity.org/2021/formulas/physics/high-school/5kpptgsvjpb0hm9hsptms0cqkyco0i2qp2.png)
![f_3 =3* (50)/(2* 2)](https://img.qammunity.org/2021/formulas/physics/high-school/2lbjsev97pwj98nqh9ctp44153sy00d001.png)
f₃ = 37.5 Hz