Question Number 228152 by fantastic2 last updated on 31/Mar/26
Commented by fantastic2 last updated on 31/Mar/26
$${a}\:{spring}\:{connects}\:{the}\:{two}\:{pole}. \\ $$$${no}\:{tension}\:{working}\:{in}\:{spring}.{now}\:{a}\:{sphere}\:{of} \\ $$$${mass}\:{m}\:{is}\:{hung}\:{from}\:{a}\:{point}\:{of}\:{the}\:{spring}. \\ $$$${and}\:{then}\:{the}\:{sphere}\:{is}\:{let}\:{to}\:{fall}.{what}\:{will} \\ $$$${be}\:{the}\:{maximum}\:{vertical}\:{displacement}. \\ $$$${spring}\:{constant}\:{k}.\:{the}\:{rope}\:{is}\:{weightless} \\ $$
Commented by mr W last updated on 01/Apr/26
$${case}\:\mathrm{1}:\:\:{the}\:{mass}\:{is}\:{released}\:{slowly} \\ $$$${case}\:\mathrm{2}:\:\:{the}\:{mass}\:{is}\:{released}\:{abruptly} \\ $$$${case}\:\mathrm{1}\:{is}\:{statical}\:{and}\:{easy}\:{to}\:{solve}. \\ $$$${case}\:\mathrm{2}\:{is}\:{dynamical}\:{and}\:{hard}\:{to}\:{solve}. \\ $$
Answered by mr W last updated on 02/Apr/26
Commented by mr W last updated on 01/Apr/26
$${case}\:\mathrm{1}:\:{the}\:{mass}\:{is}\:{released}\:{slowly}. \\ $$
Commented by mr W last updated on 01/Apr/26
$${if}\:{we}\:{have} \\ $$$${p}\:\mathrm{sin}\:\theta−{q}\:\mathrm{cos}\:\theta={r} \\ $$$${then} \\ $$$$\frac{{p}}{\:\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }}\:\mathrm{sin}\:\theta−\frac{{q}}{\:\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }}\:\mathrm{cos}\:\theta=\frac{{r}}{\:\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }} \\ $$$$\mathrm{cos}\:\varphi\:\mathrm{sin}\:\theta−\mathrm{sin}\:\varphi\:\mathrm{cos}\:\theta=\frac{{r}}{\:\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }} \\ $$$$\Rightarrow\mathrm{sin}\:\left(\theta−\varphi\right)=\frac{{r}}{\:\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }} \\ $$$${with}\:\mathrm{cos}\:\varphi=\frac{{p}}{\:\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }},\:\mathrm{sin}\:\varphi=\frac{{q}}{\:\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }} \\ $$$${or}\:\mathrm{tan}\:\varphi=\frac{{q}}{{p}},\:{i}.{e}.\:\varphi=\mathrm{tan}^{−\mathrm{1}} \frac{{q}}{{p}} \\ $$$$\Rightarrow\theta=\varphi+\mathrm{sin}^{−\mathrm{1}} \frac{{r}}{\:\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }}=\mathrm{tan}^{−\mathrm{1}} \frac{{q}}{{p}}+\mathrm{sin}^{−\mathrm{1}} \frac{{r}}{\:\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }} \\ $$
Commented by fantastic2 last updated on 01/Apr/26
$${sir}\:{i}\:{understood}\:{everything}\:{but}\:{the}\:\varphi\:{one}, \\ $$$${pls}\:{explain}\: \\ $$
Commented by fantastic2 last updated on 01/Apr/26
$${understood}\:{sir}.{thanks} \\ $$
Commented by fantastic2 last updated on 01/Apr/26
$${sir}\:{i}\:{think}\:{for}\:{the}\:\mathrm{2}{nd}\:{case}\:{we}\:{need}\:{a} \\ $$$${diff}\:{equation} \\ $$
Commented by mr W last updated on 03/Apr/26
![k_1 =(((a+b)k)/a) k_2 =(((a+b)k)/b) let λ=((mg)/((a+b)k)) (F_1 /(cos θ_2 ))=(F_2 /(cos θ_1 ))=((mg)/(sin (θ_1 +θ_2 ))) ⇒F_1 =((mg cos θ_2 )/(sin (θ_1 +θ_2 ))) ⇒F_2 =((mg cos θ_1 )/(sin (θ_1 +θ_2 ))) a_1 =a+(F_1 /k_1 )=a+((amg cos θ_2 )/((a+b)k sin (θ_1 +θ_2 )))=a[1+((λ cos θ_2 )/(sin (θ_1 +θ_2 )))] b_1 =b+(F_2 /k_2 )=b+((bmg cos θ_1 )/((a+b)k sin (θ_1 +θ_2 )))=b[1+((λ cos θ_1 )/(sin (θ_1 +θ_2 )))] (a_1 /(sin θ_2 ))=(b_1 /(sin θ_1 ))=((a+b)/(sin (θ_1 +θ_2 ))) a_1 =(((a+b) sin θ_2 )/(sin (θ_1 +θ_2 )))=a[1+((λ cos θ_2 )/(sin (θ_1 +θ_2 )))] ⇒(1+(b/a)) sin θ_2 −λ cos θ_2 =sin (θ_1 +θ_2 ) ⇒(√((1+(b/a))^2 +λ^2 ))[((1+(b/a))/( (√((1+(b/a))^2 +λ^2 )))) sin θ_2 −(λ/( (√((1+(b/a))^2 +λ^2 )))) cos θ_2 ]= sin (θ_1 +θ_2 ) ⇒(√((1+(b/a))^2 +λ^2 ))(cos ϕ_2 sin θ_2 −sin ϕ_2 cos θ_2 )= sin (θ_1 +θ_2 ) ⇒(√((1+(b/a))^2 +λ^2 )) sin (θ_2 −ϕ_2 )=sin (θ_1 +θ_2 ) ...(i) with ϕ_2 =tan^(−1) (λ/(1+(b/a))) b_1 =(((a+b) sin θ_1 )/(sin (θ_1 +θ_2 )))=b[1+((λ cos θ_1 )/(sin (θ_1 +θ_2 )))] ⇒(1+(a/b)) sin θ_1 −λ cos θ_1 =sin (θ_1 +θ_2 ) ...(i) ⇒(√((1+(a/b))^2 +λ^2 )) sin (θ_1 −ϕ_1 )=sin (θ_1 +θ_2 ) ...(ii) with ϕ_1 =tan^(−1) (λ/(1+(a/b))) (√((1+(a/b))^2 +λ^2 )) sin (θ_1 −ϕ_1 )=(√((1+(b/a))^2 +λ^2 )) sin (θ_2 −ϕ_2 )=sin (θ_1 +θ_2 )=ξ, say ⇒θ_1 =ϕ_1 +sin^(−1) (ξ/( (√((1+(a/b))^2 +λ^2 )))) ⇒θ_2 =ϕ_2 +sin^(−1) (ξ/( (√((1+(b/a))^2 +λ^2 )))) ⇒θ_1 +θ_2 =sin^(−1) ξ ⇒sin^(−1) (ξ/( (√((1+(a/b))^2 +λ^2 ))))+tan^(−1) (λ/(1+(a/b)))+sin^(−1) (ξ/( (√((1+(b/a))^2 +λ^2 ))))+tan^(−1) (λ/(1+(b/a)))=sin^(−1) ξ we can solve this equation to get ξ. a_1 =a[1+(λ/ξ) cos (ϕ_2 +sin^(−1) (ξ/( (√((1+(b/a))^2 +λ^2 )))))] b_1 =b[1+(λ/ξ) cos (ϕ_1 +sin^(−1) (ξ/( (√((1+(a/b))^2 +λ^2 )))))] h=a_1 sin θ_1 =b_1 sin θ_2 ⇒h=a[1+(λ/ξ) cos (ϕ_2 +sin^(−1) (ξ/( (√((1+(b/a))^2 +λ^2 )))))] sin (ϕ_1 +sin^(−1) (ξ/( (√((1+(a/b))^2 +λ^2 ))))) or ⇒h=b[1+(λ/ξ) cos (ϕ_1 +sin^(−1) (ξ/( (√((1+(a/b))^2 +λ^2 )))))] sin (ϕ_2 +sin^(−1) (ξ/( (√((1+(b/a))^2 +λ^2 ))))) or (a+b)h=a_1 b_1 sin (θ_1 +θ_2 ) h=((abξ)/(a+b)) (1+(λ/ξ) cos θ_1 )(1+(λ/ξ) cos θ_2 ) ⇒h=((abξ)/(a+b))[1+(λ/ξ) cos (ϕ_1 +sin^(−1) (ξ/( (√((1+(a/b))^2 +λ^2 ))))) ][1+(λ/ξ) cos (ϕ_2 +sin^(−1) (ξ/( (√((1+(b/a))^2 +λ^2 )))))] we see generally for a≠b a_1 cos θ_1 =a[1+(λ/ξ) cos (ϕ_2 +sin^(−1) (ξ/( (√((1+(b/a))^2 +λ^2 )))))] cos (ϕ_1 +sin^(−1) (ξ/( (√((1+(a/b))^2 +λ^2 ))))) ≠a b_1 cos θ_2 =b[1+(λ/ξ) cos (ϕ_1 +sin^(−1) (ξ/( (√((1+(a/b))^2 +λ^2 )))))] cos (ϕ_2 +sin^(−1) (ξ/( (√((1+(b/a))^2 +λ^2 ))))) ≠b](https://www.tinkutara.com/question/Q228170.png)
$${k}_{\mathrm{1}} =\frac{\left({a}+{b}\right){k}}{{a}} \\ $$$${k}_{\mathrm{2}} =\frac{\left({a}+{b}\right){k}}{{b}} \\ $$$${let}\:\lambda=\frac{{mg}}{\left({a}+{b}\right){k}} \\ $$$$\frac{{F}_{\mathrm{1}} }{\mathrm{cos}\:\theta_{\mathrm{2}} }=\frac{{F}_{\mathrm{2}} }{\mathrm{cos}\:\theta_{\mathrm{1}} }=\frac{{mg}}{\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} \right)} \\ $$$$\Rightarrow{F}_{\mathrm{1}} =\frac{{mg}\:\mathrm{cos}\:\theta_{\mathrm{2}} }{\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} \right)} \\ $$$$\Rightarrow{F}_{\mathrm{2}} =\frac{{mg}\:\mathrm{cos}\:\theta_{\mathrm{1}} }{\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} \right)} \\ $$$${a}_{\mathrm{1}} ={a}+\frac{{F}_{\mathrm{1}} }{{k}_{\mathrm{1}} }={a}+\frac{{amg}\:\mathrm{cos}\:\theta_{\mathrm{2}} }{\left({a}+{b}\right){k}\:\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} \right)}={a}\left[\mathrm{1}+\frac{\lambda\:\mathrm{cos}\:\theta_{\mathrm{2}} }{\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} \right)}\right] \\ $$$${b}_{\mathrm{1}} ={b}+\frac{{F}_{\mathrm{2}} }{{k}_{\mathrm{2}} }={b}+\frac{{bmg}\:\mathrm{cos}\:\theta_{\mathrm{1}} }{\left({a}+{b}\right){k}\:\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} \right)}={b}\left[\mathrm{1}+\frac{\lambda\:\mathrm{cos}\:\theta_{\mathrm{1}} }{\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} \right)}\right] \\ $$$$\frac{{a}_{\mathrm{1}} }{\mathrm{sin}\:\theta_{\mathrm{2}} }=\frac{{b}_{\mathrm{1}} }{\mathrm{sin}\:\theta_{\mathrm{1}} }=\frac{{a}+{b}}{\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} \right)} \\ $$$${a}_{\mathrm{1}} =\frac{\left({a}+{b}\right)\:\mathrm{sin}\:\theta_{\mathrm{2}} }{\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} \right)}={a}\left[\mathrm{1}+\frac{\lambda\:\mathrm{cos}\:\theta_{\mathrm{2}} }{\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} \right)}\right] \\ $$$$\Rightarrow\left(\mathrm{1}+\frac{{b}}{{a}}\right)\:\mathrm{sin}\:\theta_{\mathrm{2}} −\lambda\:\mathrm{cos}\:\theta_{\mathrm{2}} =\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} \right) \\ $$$$\Rightarrow\sqrt{\left(\mathrm{1}+\frac{{b}}{{a}}\right)^{\mathrm{2}} +\lambda^{\mathrm{2}} }\left[\frac{\mathrm{1}+\frac{{b}}{{a}}}{\:\sqrt{\left(\mathrm{1}+\frac{{b}}{{a}}\right)^{\mathrm{2}} +\lambda^{\mathrm{2}} }}\:\mathrm{sin}\:\theta_{\mathrm{2}} −\frac{\lambda}{\:\sqrt{\left(\mathrm{1}+\frac{{b}}{{a}}\right)^{\mathrm{2}} +\lambda^{\mathrm{2}} }}\:\mathrm{cos}\:\theta_{\mathrm{2}} \right]=\:\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} \right) \\ $$$$\Rightarrow\sqrt{\left(\mathrm{1}+\frac{{b}}{{a}}\right)^{\mathrm{2}} +\lambda^{\mathrm{2}} }\left(\mathrm{cos}\:\varphi_{\mathrm{2}} \:\mathrm{sin}\:\theta_{\mathrm{2}} −\mathrm{sin}\:\varphi_{\mathrm{2}} \:\mathrm{cos}\:\theta_{\mathrm{2}} \right)=\:\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} \right) \\ $$$$\Rightarrow\sqrt{\left(\mathrm{1}+\frac{{b}}{{a}}\right)^{\mathrm{2}} +\lambda^{\mathrm{2}} }\:\mathrm{sin}\:\left(\theta_{\mathrm{2}} −\varphi_{\mathrm{2}} \right)=\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} \right)\:\:\:…\left({i}\right) \\ $$$${with}\:\varphi_{\mathrm{2}} =\mathrm{tan}^{−\mathrm{1}} \frac{\lambda}{\mathrm{1}+\frac{{b}}{{a}}} \\ $$$${b}_{\mathrm{1}} =\frac{\left({a}+{b}\right)\:\mathrm{sin}\:\theta_{\mathrm{1}} }{\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} \right)}={b}\left[\mathrm{1}+\frac{\lambda\:\mathrm{cos}\:\theta_{\mathrm{1}} }{\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} \right)}\right] \\ $$$$\Rightarrow\left(\mathrm{1}+\frac{{a}}{{b}}\right)\:\mathrm{sin}\:\theta_{\mathrm{1}} −\lambda\:\mathrm{cos}\:\theta_{\mathrm{1}} =\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} \right)\:\:\:…\left({i}\right) \\ $$$$\Rightarrow\sqrt{\left(\mathrm{1}+\frac{{a}}{{b}}\right)^{\mathrm{2}} +\lambda^{\mathrm{2}} }\:\mathrm{sin}\:\left(\theta_{\mathrm{1}} −\varphi_{\mathrm{1}} \right)=\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} \right)\:\:\:…\left({ii}\right) \\ $$$${with}\:\varphi_{\mathrm{1}} =\mathrm{tan}^{−\mathrm{1}} \frac{\lambda}{\mathrm{1}+\frac{{a}}{{b}}} \\ $$$$\sqrt{\left(\mathrm{1}+\frac{{a}}{{b}}\right)^{\mathrm{2}} +\lambda^{\mathrm{2}} }\:\mathrm{sin}\:\left(\theta_{\mathrm{1}} −\varphi_{\mathrm{1}} \right)=\sqrt{\left(\mathrm{1}+\frac{{b}}{{a}}\right)^{\mathrm{2}} +\lambda^{\mathrm{2}} }\:\mathrm{sin}\:\left(\theta_{\mathrm{2}} −\varphi_{\mathrm{2}} \right)=\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} \right)=\xi,\:{say} \\ $$$$ \\ $$$$\Rightarrow\theta_{\mathrm{1}} =\varphi_{\mathrm{1}} +\mathrm{sin}^{−\mathrm{1}} \frac{\xi}{\:\sqrt{\left(\mathrm{1}+\frac{{a}}{{b}}\right)^{\mathrm{2}} +\lambda^{\mathrm{2}} }} \\ $$$$\Rightarrow\theta_{\mathrm{2}} =\varphi_{\mathrm{2}} +\mathrm{sin}^{−\mathrm{1}} \frac{\xi}{\:\sqrt{\left(\mathrm{1}+\frac{{b}}{{a}}\right)^{\mathrm{2}} +\lambda^{\mathrm{2}} }} \\ $$$$\Rightarrow\theta_{\mathrm{1}} +\theta_{\mathrm{2}} =\mathrm{sin}^{−\mathrm{1}} \xi \\ $$$$\Rightarrow\mathrm{sin}^{−\mathrm{1}} \frac{\xi}{\:\sqrt{\left(\mathrm{1}+\frac{{a}}{{b}}\right)^{\mathrm{2}} +\lambda^{\mathrm{2}} }}+\mathrm{tan}^{−\mathrm{1}} \frac{\lambda}{\mathrm{1}+\frac{{a}}{{b}}}+\mathrm{sin}^{−\mathrm{1}} \frac{\xi}{\:\sqrt{\left(\mathrm{1}+\frac{{b}}{{a}}\right)^{\mathrm{2}} +\lambda^{\mathrm{2}} }}+\mathrm{tan}^{−\mathrm{1}} \frac{\lambda}{\mathrm{1}+\frac{{b}}{{a}}}=\mathrm{sin}^{−\mathrm{1}} \xi \\ $$$${we}\:{can}\:{solve}\:{this}\:{equation}\:{to}\:{get}\:\xi. \\ $$$$ \\ $$$${a}_{\mathrm{1}} ={a}\left[\mathrm{1}+\frac{\lambda}{\xi}\:\mathrm{cos}\:\left(\varphi_{\mathrm{2}} +\mathrm{sin}^{−\mathrm{1}} \frac{\xi}{\:\sqrt{\left(\mathrm{1}+\frac{{b}}{{a}}\right)^{\mathrm{2}} +\lambda^{\mathrm{2}} }}\right)\right] \\ $$$${b}_{\mathrm{1}} ={b}\left[\mathrm{1}+\frac{\lambda}{\xi}\:\mathrm{cos}\:\left(\varphi_{\mathrm{1}} +\mathrm{sin}^{−\mathrm{1}} \frac{\xi}{\:\sqrt{\left(\mathrm{1}+\frac{{a}}{{b}}\right)^{\mathrm{2}} +\lambda^{\mathrm{2}} }}\right)\right] \\ $$$${h}={a}_{\mathrm{1}} \:\mathrm{sin}\:\theta_{\mathrm{1}} ={b}_{\mathrm{1}} \:\mathrm{sin}\:\theta_{\mathrm{2}} \\ $$$$\Rightarrow{h}={a}\left[\mathrm{1}+\frac{\lambda}{\xi}\:\mathrm{cos}\:\left(\varphi_{\mathrm{2}} +\mathrm{sin}^{−\mathrm{1}} \frac{\xi}{\:\sqrt{\left(\mathrm{1}+\frac{{b}}{{a}}\right)^{\mathrm{2}} +\lambda^{\mathrm{2}} }}\right)\right]\:\mathrm{sin}\:\left(\varphi_{\mathrm{1}} +\mathrm{sin}^{−\mathrm{1}} \frac{\xi}{\:\sqrt{\left(\mathrm{1}+\frac{{a}}{{b}}\right)^{\mathrm{2}} +\lambda^{\mathrm{2}} }}\right) \\ $$$${or} \\ $$$$\Rightarrow{h}={b}\left[\mathrm{1}+\frac{\lambda}{\xi}\:\mathrm{cos}\:\left(\varphi_{\mathrm{1}} +\mathrm{sin}^{−\mathrm{1}} \frac{\xi}{\:\sqrt{\left(\mathrm{1}+\frac{{a}}{{b}}\right)^{\mathrm{2}} +\lambda^{\mathrm{2}} }}\right)\right]\:\mathrm{sin}\:\left(\varphi_{\mathrm{2}} +\mathrm{sin}^{−\mathrm{1}} \frac{\xi}{\:\sqrt{\left(\mathrm{1}+\frac{{b}}{{a}}\right)^{\mathrm{2}} +\lambda^{\mathrm{2}} }}\right) \\ $$$${or} \\ $$$$\left({a}+{b}\right){h}={a}_{\mathrm{1}} {b}_{\mathrm{1}} \mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} \right) \\ $$$${h}=\frac{{ab}\xi}{{a}+{b}}\:\left(\mathrm{1}+\frac{\lambda}{\xi}\:\mathrm{cos}\:\theta_{\mathrm{1}} \right)\left(\mathrm{1}+\frac{\lambda}{\xi}\:\mathrm{cos}\:\theta_{\mathrm{2}} \right) \\ $$$$\Rightarrow{h}=\frac{{ab}\xi}{{a}+{b}}\left[\mathrm{1}+\frac{\lambda}{\xi}\:\mathrm{cos}\:\left(\varphi_{\mathrm{1}} +\mathrm{sin}^{−\mathrm{1}} \frac{\xi}{\:\sqrt{\left(\mathrm{1}+\frac{{a}}{{b}}\right)^{\mathrm{2}} +\lambda^{\mathrm{2}} }}\right)\:\right]\left[\mathrm{1}+\frac{\lambda}{\xi}\:\mathrm{cos}\:\left(\varphi_{\mathrm{2}} +\mathrm{sin}^{−\mathrm{1}} \frac{\xi}{\:\sqrt{\left(\mathrm{1}+\frac{{b}}{{a}}\right)^{\mathrm{2}} +\lambda^{\mathrm{2}} }}\right)\right] \\ $$$$ \\ $$$${we}\:{see}\:{generally}\:{for}\:{a}\neq{b} \\ $$$${a}_{\mathrm{1}} \:\mathrm{cos}\:\theta_{\mathrm{1}} ={a}\left[\mathrm{1}+\frac{\lambda}{\xi}\:\mathrm{cos}\:\left(\varphi_{\mathrm{2}} +\mathrm{sin}^{−\mathrm{1}} \frac{\xi}{\:\sqrt{\left(\mathrm{1}+\frac{{b}}{{a}}\right)^{\mathrm{2}} +\lambda^{\mathrm{2}} }}\right)\right]\:\mathrm{cos}\:\left(\varphi_{\mathrm{1}} +\mathrm{sin}^{−\mathrm{1}} \frac{\xi}{\:\sqrt{\left(\mathrm{1}+\frac{{a}}{{b}}\right)^{\mathrm{2}} +\lambda^{\mathrm{2}} }}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\neq{a} \\ $$$${b}_{\mathrm{1}} \:\mathrm{cos}\:\theta_{\mathrm{2}} ={b}\left[\mathrm{1}+\frac{\lambda}{\xi}\:\mathrm{cos}\:\left(\varphi_{\mathrm{1}} +\mathrm{sin}^{−\mathrm{1}} \frac{\xi}{\:\sqrt{\left(\mathrm{1}+\frac{{a}}{{b}}\right)^{\mathrm{2}} +\lambda^{\mathrm{2}} }}\right)\right]\:\mathrm{cos}\:\left(\varphi_{\mathrm{2}} +\mathrm{sin}^{−\mathrm{1}} \frac{\xi}{\:\sqrt{\left(\mathrm{1}+\frac{{b}}{{a}}\right)^{\mathrm{2}} +\lambda^{\mathrm{2}} }}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\neq{b} \\ $$
Commented by mr W last updated on 01/Apr/26
$${to}\:{be}\:{exact},\:{we}\:{get}\:{two}\:{differential} \\ $$$${equations}.\:{we}\:{can}\:{not}\:{even}\:{determine} \\ $$$${the}\:{h}_{{max}} ,\:{because}\:{at}\:{the}\:{moment}\:{a}_{{y}} =\mathrm{0}, \\ $$$${a}_{{x}} \:\neq\mathrm{0}. \\ $$$${the}\:{dynamical}\:{locus}\:{of}\:{the}\:{mass} \\ $$$${could}\:{look}\:{like}\:{this}: \\ $$
Commented by mr W last updated on 01/Apr/26
Commented by fantastic2 last updated on 01/Apr/26
$${how}\:{did}\:{you}\:{solve}\:{the}\:{equation}\:{though} \\ $$
Commented by mr W last updated on 03/Apr/26
$${if}\:{you}\:{meant}\:{this}\:{equation} \\ $$$$\mathrm{sin}^{−\mathrm{1}} \frac{\xi}{\:\sqrt{\left(\mathrm{1}+\frac{{a}}{{b}}\right)^{\mathrm{2}} +\lambda^{\mathrm{2}} }}+\mathrm{tan}^{−\mathrm{1}} \frac{\lambda}{\mathrm{1}+\frac{{a}}{{b}}}+\mathrm{sin}^{−\mathrm{1}} \frac{\xi}{\:\sqrt{\left(\mathrm{1}+\frac{{b}}{{a}}\right)^{\mathrm{2}} +\lambda^{\mathrm{2}} }}+\mathrm{tan}^{−\mathrm{1}} \frac{\lambda}{\mathrm{1}+\frac{{b}}{{a}}}=\mathrm{sin}^{−\mathrm{1}} \xi \\ $$$${it}\:{can}\:{only}\:{be}\:{solved}\:{nummerically} \\ $$$${and}\:{approximately}. \\ $$
Commented by mr W last updated on 02/Apr/26
