A string fixed at both ends is vibrating in its third harmonic. The...

The equation of the standing wave is given by $y = 0.4 \sin(0.314 x) \cos(600\pi t)$.

The amplitude of the wave at position $x$ is $A = 0.4 \sin(0.314 x)$.

For nodes, the amplitude must be zero, which gives:

$\sin(0.314 x) = 0$

$0.314 x = n\pi$, where $n = 0, 1, 2, \dots$

Using $\pi = 3.14$, we get:

$0.314 x = n(3.14)$

$x = 10n \text{ cm}$

The positions of the nodes are $x = 0, 10, 20, 30, \dots \text{ cm}$.

The wave number is $k = 0.314 \text{ cm}^{-1}$, so the wavelength is $\lambda = \dfrac{2\pi}{k} = \dfrac{2 \times 3.14}{0.314} = 20 \text{ cm}$.

Since the string is vibrating in its third harmonic, the length of the string is $L = 3 \left(\dfrac{\lambda}{2}\right) = 3 \times \dfrac{20}{2} = 30 \text{ cm}$.

Thus, the nodes on the string are located at $x = 0, 10 \text{ cm}, 20 \text{ cm}, 30 \text{ cm}$.