Tension in the rod

First, lets find $\lambda_o$

We know, $M=\displaystyle\int_{0}^{L} \lambda(x) dx$

So, $M=\displaystyle\int_{0}^{L} \lambda_o \sin \left(\dfrac{\pi x}{L}\right) dx$

$\rightarrow M=\dfrac{2L \lambda_o}{\pi}$

$\rightarrow \lambda_o=\dfrac{M \pi}{2L}$

According to the question, $M=1 kg$ . So,

$\lambda_o=\dfrac{\pi}{2L}$

Now that we know $\lambda_o$ , we will equate the force due to the part of the rope that is below $x=\frac{3L}{4}$ to the tension.

The force due to that part of the rope will be the mass of that part times $g$ .

So,

$T=g\displaystyle\int_{\frac{3L}{4}}^{L} \lambda(x) dx$

$=\dfrac{g(\sqrt{2}-1)}{2\sqrt{2}} \approx \boxed{1.435}$

In order to find the tension, we must equate the gravity acting on the rod below a point:

$T(\ell) \ = \ \displaystyle\int_{\ell}^{L} \ g \lambda(x) \ dx \ = \ g\lambda_0 \ \displaystyle\int_{\ell}^{L} \ \sin \Big( \frac{\pi x}{L} \Big) \ dx \ = \ g\lambda_0 \Big( -\cos(\pi) \ + \ \cos(\frac{\pi \ell}{L}) \Big)$

So now we need to find $\lambda_0$ . We know that:

$m \ = \ \displaystyle\int_{0}^{L} \ \lambda_0 \ \sin \Big( \frac{\pi x}{L} \Big) \ dx \ \Rightarrow \ \lambda_0 \ = \ \frac{\pi}{2 L}$

Thus we have:

$T(\frac{3 L}{4}) \ = \ \frac{g \pi}{2L} \ \Big( 1 \ + \ \cos(\frac{3 \pi}{4}) \Big) \ = \ 1.425176772 \ N$