Answer
$12.27\;N$
Work Step by Step
let, the position of a point on the string after time $t$ makes the angle $\theta$ with
the positive $x$ axis.
The tension $(T)$ is along the string and its direction at that point also makes the angle $\theta$ with the positive $x$ axis.
Therefore, the transverse component of the tension is:
$T_{trans}=T\sin\theta$
The displacement of any point on the string follows the following equation
$y(x,t)=y_m\sin(kx-\omega t)$
Now,
$\tan\theta=\frac{\partial y}{\partial x} $
or, $\tan\theta=ky_m\cos(kx-\omega t)$
Therefore,
$\tan\theta|_{max}=ky_m$
or, $\theta_{max}=\tan^{-1}(ky_m)$
According to the given values
$2y_m=1\times10^{-2}\;m$
or, $y_m=5\times10^{-3}\;m$
$f=120\;Hz$
$\omega=2\pi f$
or, $\omega=240\pi\;rad/s$
$v=\sqrt {\frac{T}{\mu}}$
or, $v=\sqrt {\frac{90}{120\times10^{-3}}}\; m/s$
or, or, $v=5\sqrt {30}\; m/s$
We know
$v=\frac{\omega}{k}$
or, $k=\frac{\omega}{v}$
or, $k=\frac{240\pi}{5\sqrt {30}}\;rad/m$
or, $k=27.53\;rad/m$
Substituting the above values, we obtain
$\theta_{max}=\tan^{-1}(27.53\times5\times10^{-3})\;rad$
or, $\theta_{max}=0.13679\;rad$
Therefore, the maximum value of the transverse component of the tension is:
$T^{max}_{trans}=T\sin\theta_{max}$
or, $T^{max}_{trans}=90\sin(0.13679)\;N$
or, $T^{max}_{trans}=12.27\;N$
Thus, the maximum value of the transverse component of the tension is $12.27\;N$