91Ó°ÊÓ

A simple harmonic oscillator, of mass \(m\) and natural frequency \(\omega_{0}\), experiences an oscillating driving force \(f(t)=m a \cos \omega t .\) Therefore, its equation of motion is $$ \frac{d^{2} x}{d t^{2}}+\omega_{0}^{2} x=a \cos \omega t $$ where \(x\) is its position. Given that at \(t=0\) we have \(x=d x / d t=0\), find the function \(x(t) .\) Describe the solution if \(\omega\) is approximately, but not exactly, equal to \(\omega_{0}\).

The theory of bent beams shows that at any point in the beam the 'bending moment' is given by \(K / \rho\), where \(K\) is a constant (that depends upon the beam material and cross-sectional shape) and \(\rho\) is the radius of curvature at that point. Consider a light beam of length \(L\) whose ends, \(x=0\) and \(x=L\), are supported at the same vertical height and which has a weight \(W\) suspended from its centre. Verify that at any point \(x(0 \leq x \leq L / 2\) for definiteness) the net magnitude of the bending moment (bending moment \(=\) force \(\times\) perpendicular distance) due to the weight and support reactions, evaluated on either side of \(x\), is \(W x / 2\). If the beam is only slightly bent, so that \((d y / d x)^{2} \ll 1\), where \(y=y(x)\) is the downward displacement of the beam at \(x\), show that the beam profile satisfies the approximate equation $$ \frac{d^{2} y}{d x^{2}}=-\frac{W x}{2 K} $$ By integrating this equation twice and using physically imposed conditions on your solution at \(x=0\) and \(x=L / 2\), show that the downward displacement at the centre of the beam is \(W L^{3} /(48 K)\).