A 2.0-m-long string vibrates at its second-harmonic frequency with a maximum amplitude of 2.0 cm. One end of the string is at x = 0 cm. Find the oscillation amplitude at x = 10, 20, 30, 40, and 50 cm.

Length of string L= 2m
Second harmonic frequency, n= 2
The amplitude of the wave= 2 cm = 0.02 m

Let the wavelength is denoted by λ

The wavelength of the nth harmonic frequency is given by
${\lambda }_n=\frac{2L}{n}$

Here n=2,

$$\begin{gathered} {\lambda }_2=\frac{2L}{2}=L \\ {\lambda }_2=2m \end{gathered}$$
Thus, the wavelength of the wave is 2m

The general expression of the amplitude of a wave is given by
$A\left(x\right)=\left(2a\right)\sin \left(\frac{2\pi x}{\lambda }\right)$

Substitute the known values

$$\begin{gathered} A\left(x\right)=\left(0.02\right)\sin \left(\frac{2\pi x}{2}\right) \\ A\left(x\right)=\left(0.02\right)\sin \left(\pi x\right)......\left(1\right) \end{gathered}$$

Substitute x= 10 cm=0.1m in equation (1)
$A(0.1)=(0.02)\sin (\pi (0.1\left)\right)=6.2mm$

Put x= 20cm = 0.2m in equation (1)
$A(0.2)=(0.02)\sin (\pi (0.2\left)\right)=11.8mm$

Put x= 30cm = 0.3m in equation (1)
$A(0.3)=(0.02)\sin (\pi (0.3\left)\right)=16.2mm$

Put x= 40cm = 0.4m in equation (1)
$A(0.4)=(0.02)\sin (\pi (0.4\left)\right)=19.0mm$

Put x= 50cm = 0.5m in equation (1)
$A(0.5)=(0.02)\sin (\pi (0.5\left)\right)=20.0mm$

Thus, the amplitude of the wave at x =10 cm, 20cm, 30cm, 40cm and 50cm is 6.2mm, 11.8mm, 16.2 mm, 19.0 mm, 19.0 mm and 20mm respectively