A sinusoidally varying force is applied at one end \(\left(x=x_{1}=0\right)\) of a long slender rod, terminated by its characteristic impedance at the other end \((x=l)\) (see Sec. 4.2). Show that the transfer function for the force at any intermediate position \(x_{2}\) is \(T(\omega)=e^{-j \omega t_{2}}\), where \(t_{2}=x_{2} / c\), with \(c\) the wave velocity. If a force \(f(t)\) is applied at \(x_{1}=0\), show that the force function at \(x_{2}\) is \(f\left(t-t_{2}\right)\).

To find the time delay \(t_{2}\) at the intermediate position \(x_{2}\), we will make use of the given relation \(t_{2} = x_{2}/c\), where \(c\) is the wave velocity. This relationship is derived from the fact that the time taken for a wave to travel a distance \(x_{2}\) through the rod is proportional to the distance and inversely proportional to the wave velocity.

The sinusoidally varying force applied at \(x_{1} = 0\) can be described as a complex exponential of the form \(F(\omega) = e^{j \omega t}\), where \(\omega\) is the angular frequency. Due to the structure of the problem, the complex exponential solution is preserved along the length of the rod with the wave velocity \(c\). The force at \(x_{2}\) can be described as \(F(\omega)e^{-j \omega t_{2}}\), which includes the phase shift introduced by the time delay \(t_{2}\). Therefore, the transfer function \(T(\omega)\) is given by: $$T(\omega) = e^{-j \omega t_{2}}$$

Given the force function at \(x_{1}=0\) as \(f(t)\), it can be expressed in the frequency domain as \(F(\omega)\) using the Fourier transform. We can use the transfer function \(T(\omega)\) to find the force function at \(x_{2}\) in the frequency domain: $$F_{x_{2}}(\omega) = F(\omega)T(\omega) = F(\omega) e^{-j \omega t_{2}}$$ To find the force function at \(x_{2}\) in the time domain, we take the inverse Fourier transform of \(F_{x_{2}}(\omega)\): $$f_{x_{2}}(t) = \mathcal{F}^{-1}\{F_{x_{2}}(\omega)\} = \mathcal{F}^{-1}\{F(\omega)e^{-j \omega t_{2}}\}$$ By the time-shifting property of the inverse Fourier transform: $$f_{x_{2}}(t) = f(t-t_{2})$$ So, the force function at \(x_{2}\) is \(f(t-t_{2})\), as required.