Lengthening Pendulum Problem #19 : I think I have solved it. Angular momentum is conserved according to eq 3 on p 268 of the following article : audiophile.tam.cornell.edu/…. So at the lowest point we can write Lv0=(L+δ)v1Lv0=(L+δ)v1 or $v_0^2=(1+x)^2v_1^2$ wherewhere $v_0, v_1$ are the velocities to left and right of the lowest point and are the velocities to left and right of the lowest point and $x=\frac{\delta}{L}$.

On the LHS of the swing the mass falls a distance h0=L(1−cosθ0)h0=L(1−cos⁡θ0) so that $v_0^2=2gh_0v0^2=2gh_0$. On the RHS of the swing the mass rises through a height of h1…

Lengthening Pendulum Problem #19 : I think I have solved it. Angular momentum is conserved according to eq 3 on p 268 of the following article : audiophile.tam.cornell.edu/…. So at the lowest point we can write Lv0=(L+δ)v1Lv0=(L+δ)v1 or $v_0^2=(1+x)^2v_1^2$ wherewhere $v_0, v_1$ are the velocities to left and right of the lowest point and are the velocities to left and right of the lowest point and $x=\frac{\delta}{L}$.

On the LHS of the swing the mass falls a distance h0=L(1−cosθ0)h0=L(1−cos⁡θ0) so that $v_0^2=2gh_0v0^2=2gh_0$. On the RHS of the swing the mass rises through a height of h1…

@Nobodyrecognizeable Lengthening Pendulum Problem #19 : I think I have solved it. Angular momentum is conserved according to eq 3 on p 268 of the following article : audiophile.tam.cornell.edu/randpdf/swing.pdf. So at the lowest point we can write $Lv_0=(L+\delta)v_1$ or v_0^2=(1+x)^2v_1^2$ where $v_0, v_1$ are the velocities to left and right of the lowest point and $x=\frac{\delta}{L}$.
On the LHS of the swing the mass falls a distance $h_0=L(1-\cos\theta_0)$ so that $v_0^2=2gh_0$. On the RHS of the swing the mass rises through a height of $h_1=(L+\delta)(1-\cos\theta_1)$ when it reaches …