Menu Close

Question-225563

Tinku Tara 0 Comments
Pin




Question Number 225563 by mr W last updated on 03/Nov/25
Commented by mr W last updated on 03/Nov/25
Commented by fantastic last updated on 03/Nov/25
gonna do the same q when μ=ζx
$${gonna}\:{do}\:{the}\:{same}\:{q} \\ $$$${when}\:\mu=\zeta{x} \\ $$
Commented by fantastic last updated on 03/Nov/25
I got λ=((16Mg)/(3k))
$${I}\:{got}\:\lambda=\frac{\mathrm{16}{Mg}}{\mathrm{3}{k}} \\ $$
Commented by mr W last updated on 03/Nov/25
x_(max) =((16Mg)/(3k)) is right.
$${x}_{{max}} =\frac{\mathrm{16}{Mg}}{\mathrm{3}{k}}\:{is}\:{right}. \\ $$
Answered by mr W last updated on 04/Nov/25
v=(dx/dt) a_1 =(dv/dt)=v(dv/dx) a_3 =a_1 +a_r a_2 =a_1 −a_r 2M(a_1 +a_r )=2Mg−T_2 ...(i) M(a_1 −a_r )=Mg−T_2 ...(ii) (i)/2+(ii): 2Ma_1 =Mg−(T_2 /2)+Mg−T_2 ⇒T_2 =((4M(g−a_1 ))/3) T_1 =2T_2 2Ma_1 =T_1 −kx 2Ma_1 =((8M(g−a_1 ))/3)−kx 14Ma_1 =8Mg−3kx 14M×v(dv/dx)=8Mg−3kx 14Mvdv=(8Mg−3kx)dx 14M∫_0 ^v vdv=∫_0 ^x (8Mg−3kx)dx Mv^2 =(8Mg−((3kx)/2))x v^2 =(8g−((3kx)/(2M)))x v=(√(x(8g−((3kx)/(2M))))) v=0 at x_(max) : (√(x_(max) (8g−((3kx_(max) )/(2M)))))=0 ⇒x_(max) =((16Mg)/(3k)) v=(√(((2M)/(3k))×((3kx)/(2M))(8g−((3kx)/(2M))))) v_(max) is when ((3kx)/(2M))=8g−((3kx)/(2M)), i.e. when x=((8Mg)/(3k))=(x_(max) /2). v_(max) =(√(((2M)/(3k))×(4g)^2 ))=4g(√((2M)/(3k))) v=(dx/dt)=(√(((3k)/(2M))(((16Mg)/(3k))−x)x)) t=(√((2M)/(3k)))∫_0 ^x (dx/( (√((((16Mg)/(3k))−x)x)))) ⇒t=2(√((2M)/(3k))) sin^(−1) (√((3kx)/(16Mg))) x=x_(max) : T=π(√((2M)/(3k)))
$${v}=\frac{{dx}}{{dt}} \\ $$$${a}_{\mathrm{1}} =\frac{{dv}}{{dt}}={v}\frac{{dv}}{{dx}} \\ $$$${a}_{\mathrm{3}} ={a}_{\mathrm{1}} +{a}_{{r}} \\ $$$${a}_{\mathrm{2}} ={a}_{\mathrm{1}} −{a}_{{r}} \\ $$$$\mathrm{2}{M}\left({a}_{\mathrm{1}} +{a}_{{r}} \right)=\mathrm{2}{Mg}−{T}_{\mathrm{2}} \:\:\:…\left({i}\right) \\ $$$${M}\left({a}_{\mathrm{1}} −{a}_{{r}} \right)={Mg}−{T}_{\mathrm{2}} \:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)/\mathrm{2}+\left({ii}\right): \\ $$$$\mathrm{2}{Ma}_{\mathrm{1}} ={Mg}−\frac{{T}_{\mathrm{2}} }{\mathrm{2}}+{Mg}−{T}_{\mathrm{2}} \\ $$$$\Rightarrow{T}_{\mathrm{2}} =\frac{\mathrm{4}{M}\left({g}−{a}_{\mathrm{1}} \right)}{\mathrm{3}} \\ $$$${T}_{\mathrm{1}} =\mathrm{2}{T}_{\mathrm{2}} \\ $$$$\mathrm{2}{Ma}_{\mathrm{1}} ={T}_{\mathrm{1}} −{kx} \\ $$$$\mathrm{2}{Ma}_{\mathrm{1}} =\frac{\mathrm{8}{M}\left({g}−{a}_{\mathrm{1}} \right)}{\mathrm{3}}−{kx} \\ $$$$\mathrm{14}{Ma}_{\mathrm{1}} =\mathrm{8}{Mg}−\mathrm{3}{kx} \\ $$$$\mathrm{14}{M}×{v}\frac{{dv}}{{dx}}=\mathrm{8}{Mg}−\mathrm{3}{kx} \\ $$$$\mathrm{14}{Mvdv}=\left(\mathrm{8}{Mg}−\mathrm{3}{kx}\right){dx} \\ $$$$\mathrm{14}{M}\int_{\mathrm{0}} ^{{v}} {vdv}=\int_{\mathrm{0}} ^{{x}} \left(\mathrm{8}{Mg}−\mathrm{3}{kx}\right){dx} \\ $$$${Mv}^{\mathrm{2}} =\left(\mathrm{8}{Mg}−\frac{\mathrm{3}{kx}}{\mathrm{2}}\right){x} \\ $$$${v}^{\mathrm{2}} =\left(\mathrm{8}{g}−\frac{\mathrm{3}{kx}}{\mathrm{2}{M}}\right){x} \\ $$$${v}=\sqrt{{x}\left(\mathrm{8}{g}−\frac{\mathrm{3}{kx}}{\mathrm{2}{M}}\right)} \\ $$$${v}=\mathrm{0}\:{at}\:{x}_{{max}} : \\ $$$$\sqrt{{x}_{{max}} \left(\mathrm{8}{g}−\frac{\mathrm{3}{kx}_{{max}} }{\mathrm{2}{M}}\right)}=\mathrm{0} \\ $$$$\Rightarrow{x}_{{max}} =\frac{\mathrm{16}{Mg}}{\mathrm{3}{k}} \\ $$$${v}=\sqrt{\frac{\mathrm{2}{M}}{\mathrm{3}{k}}×\frac{\mathrm{3}{kx}}{\mathrm{2}{M}}\left(\mathrm{8}{g}−\frac{\mathrm{3}{kx}}{\mathrm{2}{M}}\right)} \\ $$$${v}_{{max}} \:{is}\:{when}\:\frac{\mathrm{3}{kx}}{\mathrm{2}{M}}=\mathrm{8}{g}−\frac{\mathrm{3}{kx}}{\mathrm{2}{M}},\:{i}.{e}. \\ $$$${when}\:{x}=\frac{\mathrm{8}{Mg}}{\mathrm{3}{k}}=\frac{{x}_{{max}} }{\mathrm{2}}. \\ $$$${v}_{{max}} =\sqrt{\frac{\mathrm{2}{M}}{\mathrm{3}{k}}×\left(\mathrm{4}{g}\right)^{\mathrm{2}} }=\mathrm{4}{g}\sqrt{\frac{\mathrm{2}{M}}{\mathrm{3}{k}}} \\ $$$${v}=\frac{{dx}}{{dt}}=\sqrt{\frac{\mathrm{3}{k}}{\mathrm{2}{M}}\left(\frac{\mathrm{16}{Mg}}{\mathrm{3}{k}}−{x}\right){x}} \\ $$$${t}=\sqrt{\frac{\mathrm{2}{M}}{\mathrm{3}{k}}}\int_{\mathrm{0}} ^{{x}} \frac{{dx}}{\:\sqrt{\left(\frac{\mathrm{16}{Mg}}{\mathrm{3}{k}}−{x}\right){x}}} \\ $$$$\Rightarrow{t}=\mathrm{2}\sqrt{\frac{\mathrm{2}{M}}{\mathrm{3}{k}}}\:\mathrm{sin}^{−\mathrm{1}} \sqrt{\frac{\mathrm{3}{kx}}{\mathrm{16}{Mg}}} \\ $$$${x}={x}_{{max}} :\:{T}=\pi\sqrt{\frac{\mathrm{2}{M}}{\mathrm{3}{k}}} \\ $$
Commented by fantastic last updated on 03/Nov/25
got it t=∫_0 ^x_(max) (dx/( (√(2a_1 x−x^2 (ξg+(k/M_1 ))))))
$${got}\:{it}\:{t}=\int_{\mathrm{0}} ^{{x}_{{max}} } \frac{{dx}}{\:\sqrt{\mathrm{2}{a}_{\mathrm{1}} {x}−{x}^{\mathrm{2}} \left(\xi{g}+\frac{{k}}{{M}_{\mathrm{1}} }\right)}} \\ $$
Commented by fantastic last updated on 04/Nov/25
wow sir
$${wow}\:{sir} \\ $$
Answered by fantastic last updated on 03/Nov/25
Commented by fantastic last updated on 03/Nov/25
k⇒δ ,x⇒λ T_1 =2T_2 M_1 a_1 =2T_2 T_2 =((M_1 a_1 )/2) from M_2 &M_3 M_3 g−T_2 =M_3 a ...i T_2 −M_2 g=M_2 a ...ii i+ii g(M_3 −M_2 )=a(M_3 +M_2 ) a=(((M_3 −M_2 )g)/((M_3 +M_2 ))) T_2 =M_2 (g+a) =M_2 g(1+(((M_3 −M_2 ))/((M_3 +M_2 )))) =M_2 g((2M_3 )/(M_3 +M_2 )) ∴((M_1 a_1 )/2)=M_2 g((2M_3 )/(M_3 +M_2 )) ⇒a_1 =((4M_2 M_3 g)/(M_1 (M_2 +M_3 ))) M_1 is going → so friction and spring force will act← total force acting against M_1 μn+δλ=ξλM_1 g+δλ backward acc=ξλg+((δλ)/M_1 ) relative acc=a_1 −ξλg−((δλ)/M_1 ) =((4M_2 M_3 g)/(M_1 (M_2 +M_3 )))−ξλg−((δλ)/M_1 ) so v(dv/dλ)=((4M_2 M_3 g)/(M_1 (M_2 +M_3 )))−ξλg−((δλ)/M_1 ) vdv=((4M_2 M_3 g)/(M_1 (M_2 +M_3 )))dλ−ξλgdλ−((δλ)/M_1 )dλ ∫_0 ^v vdv=∫_0 ^λ ((4M_2 M_3 g)/(M_1 (M_2 +M_3 )))dλ−∫_0 ^λ ξλgdλ−∫_0 ^λ ((δλ)/M_1 )dλ (v^2 /2)=((4M_2 M_3 g)/(M_1 (M_2 +M_3 )))λ−((ξλ^2 g)/2)−((δλ^2 )/(2M_1 )) .....(I) at λ_(max) v=0 so ((4M_2 M_3 g)/(M_1 (M_2 +M_3 )))λ=((ξλ^2 g)/2)+((δλ^2 )/(2M_1 )) ⇒λ_(max) =(((8M_2 M_3 g)/(M_1 (M_3 +M_2 )))/((ξgM_1 +δ)/M_1 ))=((8M_2 M_3 g)/((ξgM_1 +δ)(M_3 +M_2 ))) example M_1 =2M ,M_2 =M,M_3 =2M μ=0 λ_(max) =((8×M×2M×g)/((0+δ)(2M+M)))=((16Mg)/(3δ))✓ from (I) v=(√(2a_1 λ−λ^2 (ξg+(δ/M_1 )))) (dλ/dt)=(√(2a_1 λ−λ^2 (ξg+(δ/M_1 )))) dt=(dλ/( (√(2a_1 λ−λ^2 (ξg+(δ/M_1 )))))) ∫_0 ^t dt=∫_0 ^λ_(max) (dλ/( (√(2a_1 λ−λ^2 (ξg+(δ/M_1 )))))) let (ξg+(δ/M_1 ))=ϕ ⇒(√(2a_1 λ−λ^2 ϕ))=(√ϕ)×(√(((2a_1 )/ϕ)λ−λ^2 )) let λ=((2a_1 )/ϕ)sin^2 θ dλ=((4a_1 )/ϕ)sin θcos θ dθ (√(((2a_1 )/ϕ)λ−λ^2 ))=((2a_1 )/ϕ)sin θcos θ ∫_0 ^λ_(max) (dλ/( (√(2a_1 λ−λ^2 (ξg+(δ/M_1 )))))) =(1/( (√ϕ)))∫_0 ^λ_(max) ((((4a_1 )/ϕ)sin θcos θdθ)/(((2a_1 )/ϕ)sin θcos θ))=(1/( (√ϕ)))∫_0 ^λ_(max) 2dθ=(2/( (√ϕ)))[θ]_0 ^λ_(max) λ=((2a_1 )/ϕ)sin^2 θ sin θ=(√((λϕ)/(2a_1 )))⇒θ=sin^(−1) ((√((λϕ)/(2a_1 )))) ⇒(2/( (√ϕ)))[sin^(−1) ((√((λϕ)/(2a_1 ))))]_0 ^λ_(max) t=(2/( (√((ξg+(δ/M_1 ))))))sin^(−1) ((√(((((8M_2 M_3 g)/((ξgM_1 +δ)(M_3 +M_2 ))))(ξg+(δ/M_1 )))/(2((4M_2 M_3 g)/(M_1 (M_2 +M_3 ))))))) M_1 =2M ,M_2 =M,M_3 =2M μ=0 t=(2/( (√(δ/(2M)))))sin^(−1) ((√((((16Mg)/(3δ))×(δ/(2M)))/(2×(4/3)g)))) =((2(√(2M)))/( (√δ)))sin^(−1) (1)=(π/2)2(√((2M)/δ)) So t=π(√((2M)/δ)) when λ=λ_(max)
$${k}\Rightarrow\delta\:,{x}\Rightarrow\lambda \\ $$$${T}_{\mathrm{1}} =\mathrm{2}{T}_{\mathrm{2}} \\ $$$${M}_{\mathrm{1}} {a}_{\mathrm{1}} =\mathrm{2}{T}_{\mathrm{2}} \\ $$$${T}_{\mathrm{2}} =\frac{{M}_{\mathrm{1}} {a}_{\mathrm{1}} }{\mathrm{2}} \\ $$$${from}\:{M}_{\mathrm{2}} \:\&{M}_{\mathrm{3}} \\ $$$${M}_{\mathrm{3}} {g}−{T}_{\mathrm{2}} ={M}_{\mathrm{3}} {a}\:\:\:…{i} \\ $$$${T}_{\mathrm{2}} −{M}_{\mathrm{2}} {g}={M}_{\mathrm{2}} {a}\:\:\:…{ii} \\ $$$${i}+{ii} \\ $$$${g}\left({M}_{\mathrm{3}} −{M}_{\mathrm{2}} \right)={a}\left({M}_{\mathrm{3}} +{M}_{\mathrm{2}} \right) \\ $$$${a}=\frac{\left({M}_{\mathrm{3}} −{M}_{\mathrm{2}} \right){g}}{\left({M}_{\mathrm{3}} +{M}_{\mathrm{2}} \right)} \\ $$$${T}_{\mathrm{2}} ={M}_{\mathrm{2}} \left({g}+{a}\right) \\ $$$$={M}_{\mathrm{2}} {g}\left(\mathrm{1}+\frac{\left({M}_{\mathrm{3}} −{M}_{\mathrm{2}} \right)}{\left({M}_{\mathrm{3}} +{M}_{\mathrm{2}} \right)}\right) \\ $$$$={M}_{\mathrm{2}} {g}\frac{\mathrm{2}{M}_{\mathrm{3}} }{{M}_{\mathrm{3}} +{M}_{\mathrm{2}} } \\ $$$$\therefore\frac{{M}_{\mathrm{1}} {a}_{\mathrm{1}} }{\mathrm{2}}={M}_{\mathrm{2}} {g}\frac{\mathrm{2}{M}_{\mathrm{3}} }{{M}_{\mathrm{3}} +{M}_{\mathrm{2}} } \\ $$$$\Rightarrow{a}_{\mathrm{1}} =\frac{\mathrm{4}{M}_{\mathrm{2}} {M}_{\mathrm{3}} {g}}{{M}_{\mathrm{1}} \left({M}_{\mathrm{2}} +{M}_{\mathrm{3}} \right)} \\ $$$${M}_{\mathrm{1}} \:{is}\:{going}\:\rightarrow \\ $$$${so}\:{friction}\:{and}\:{spring}\:{force}\:{will}\:{act}\leftarrow \\ $$$${total}\:{force}\:{acting}\:{against}\:{M}_{\mathrm{1}} \\ $$$$\mu{n}+\delta\lambda=\xi\lambda{M}_{\mathrm{1}} {g}+\delta\lambda \\ $$$${backward}\:{acc}=\xi\lambda{g}+\frac{\delta\lambda}{{M}_{\mathrm{1}} } \\ $$$${relative}\:{acc}={a}_{\mathrm{1}} −\xi\lambda{g}−\frac{\delta\lambda}{{M}_{\mathrm{1}} } \\ $$$$=\frac{\mathrm{4}{M}_{\mathrm{2}} {M}_{\mathrm{3}} {g}}{{M}_{\mathrm{1}} \left({M}_{\mathrm{2}} +{M}_{\mathrm{3}} \right)}−\xi\lambda{g}−\frac{\delta\lambda}{{M}_{\mathrm{1}} } \\ $$$${so} \\ $$$${v}\frac{{dv}}{{d}\lambda}=\frac{\mathrm{4}{M}_{\mathrm{2}} {M}_{\mathrm{3}} {g}}{{M}_{\mathrm{1}} \left({M}_{\mathrm{2}} +{M}_{\mathrm{3}} \right)}−\xi\lambda{g}−\frac{\delta\lambda}{{M}_{\mathrm{1}} } \\ $$$${vdv}=\frac{\mathrm{4}{M}_{\mathrm{2}} {M}_{\mathrm{3}} {g}}{{M}_{\mathrm{1}} \left({M}_{\mathrm{2}} +{M}_{\mathrm{3}} \right)}{d}\lambda−\xi\lambda{gd}\lambda−\frac{\delta\lambda}{{M}_{\mathrm{1}} }{d}\lambda \\ $$$$\int_{\mathrm{0}} ^{{v}} {vdv}=\int_{\mathrm{0}} ^{\lambda} \frac{\mathrm{4}{M}_{\mathrm{2}} {M}_{\mathrm{3}} {g}}{{M}_{\mathrm{1}} \left({M}_{\mathrm{2}} +{M}_{\mathrm{3}} \right)}{d}\lambda−\int_{\mathrm{0}} ^{\lambda} \xi\lambda{gd}\lambda−\int_{\mathrm{0}} ^{\lambda} \frac{\delta\lambda}{{M}_{\mathrm{1}} }{d}\lambda \\ $$$$\frac{{v}^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{4}{M}_{\mathrm{2}} {M}_{\mathrm{3}} {g}}{{M}_{\mathrm{1}} \left({M}_{\mathrm{2}} +{M}_{\mathrm{3}} \right)}\lambda−\frac{\xi\lambda^{\mathrm{2}} {g}}{\mathrm{2}}−\frac{\delta\lambda^{\mathrm{2}} }{\mathrm{2}{M}_{\mathrm{1}} }\:…..\left({I}\right) \\ $$$${at}\:\lambda_{{max}} \:{v}=\mathrm{0} \\ $$$${so}\:\frac{\mathrm{4}{M}_{\mathrm{2}} {M}_{\mathrm{3}} {g}}{{M}_{\mathrm{1}} \left({M}_{\mathrm{2}} +{M}_{\mathrm{3}} \right)}\lambda=\frac{\xi\lambda^{\mathrm{2}} {g}}{\mathrm{2}}+\frac{\delta\lambda^{\mathrm{2}} }{\mathrm{2}{M}_{\mathrm{1}} } \\ $$$$\Rightarrow\lambda_{{max}} =\frac{\frac{\mathrm{8}{M}_{\mathrm{2}} {M}_{\mathrm{3}} {g}}{{M}_{\mathrm{1}} \left({M}_{\mathrm{3}} +{M}_{\mathrm{2}} \right)}}{\frac{\xi{gM}_{\mathrm{1}} +\delta}{{M}_{\mathrm{1}} }}=\frac{\mathrm{8}{M}_{\mathrm{2}} {M}_{\mathrm{3}} {g}}{\left(\xi{gM}_{\mathrm{1}} +\delta\right)\left({M}_{\mathrm{3}} +{M}_{\mathrm{2}} \right)} \\ $$$${example} \\ $$$${M}_{\mathrm{1}} =\mathrm{2}{M}\:,{M}_{\mathrm{2}} ={M},{M}_{\mathrm{3}} =\mathrm{2}{M}\:\mu=\mathrm{0} \\ $$$$\lambda_{{max}} =\frac{\mathrm{8}×{M}×\mathrm{2}{M}×{g}}{\left(\mathrm{0}+\delta\right)\left(\mathrm{2}{M}+{M}\right)}=\frac{\mathrm{16}{Mg}}{\mathrm{3}\delta}\checkmark \\ $$$${from}\:\left({I}\right) \\ $$$${v}=\sqrt{\mathrm{2}{a}_{\mathrm{1}} \lambda−\lambda^{\mathrm{2}} \left(\xi{g}+\frac{\delta}{{M}_{\mathrm{1}} }\right)} \\ $$$$\frac{{d}\lambda}{{dt}}=\sqrt{\mathrm{2}{a}_{\mathrm{1}} \lambda−\lambda^{\mathrm{2}} \left(\xi{g}+\frac{\delta}{{M}_{\mathrm{1}} }\right)} \\ $$$${dt}=\frac{{d}\lambda}{\:\sqrt{\mathrm{2}{a}_{\mathrm{1}} \lambda−\lambda^{\mathrm{2}} \left(\xi{g}+\frac{\delta}{{M}_{\mathrm{1}} }\right)}} \\ $$$$\int_{\mathrm{0}} ^{{t}} {dt}=\int_{\mathrm{0}} ^{\lambda_{{max}} } \frac{{d}\lambda}{\:\sqrt{\mathrm{2}{a}_{\mathrm{1}} \lambda−\lambda^{\mathrm{2}} \left(\xi{g}+\frac{\delta}{{M}_{\mathrm{1}} }\right)}} \\ $$$${let}\:\left(\xi{g}+\frac{\delta}{{M}_{\mathrm{1}} }\right)=\varphi \\ $$$$\Rightarrow\sqrt{\mathrm{2}{a}_{\mathrm{1}} \lambda−\lambda^{\mathrm{2}} \varphi}=\sqrt{\varphi}×\sqrt{\frac{\mathrm{2}{a}_{\mathrm{1}} }{\varphi}\lambda−\lambda^{\mathrm{2}} } \\ $$$${let}\:\lambda=\frac{\mathrm{2}{a}_{\mathrm{1}} }{\varphi}\mathrm{sin}\:^{\mathrm{2}} \theta \\ $$$${d}\lambda=\frac{\mathrm{4}{a}_{\mathrm{1}} }{\varphi}\mathrm{sin}\:\theta\mathrm{cos}\:\theta\:{d}\theta \\ $$$$\sqrt{\frac{\mathrm{2}{a}_{\mathrm{1}} }{\varphi}\lambda−\lambda^{\mathrm{2}} }=\frac{\mathrm{2}{a}_{\mathrm{1}} }{\varphi}\mathrm{sin}\:\theta\mathrm{cos}\:\theta \\ $$$$\int_{\mathrm{0}} ^{\lambda_{{max}} } \frac{{d}\lambda}{\:\sqrt{\mathrm{2}{a}_{\mathrm{1}} \lambda−\lambda^{\mathrm{2}} \left(\xi{g}+\frac{\delta}{{M}_{\mathrm{1}} }\right)}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\varphi}}\int_{\mathrm{0}} ^{\lambda_{{max}} } \frac{\frac{\mathrm{4}{a}_{\mathrm{1}} }{\varphi}\mathrm{sin}\:\theta\mathrm{cos}\:\theta{d}\theta}{\frac{\mathrm{2}{a}_{\mathrm{1}} }{\varphi}\mathrm{sin}\:\theta\mathrm{cos}\:\theta}=\frac{\mathrm{1}}{\:\sqrt{\varphi}}\int_{\mathrm{0}} ^{\lambda_{{max}} } \mathrm{2}{d}\theta=\frac{\mathrm{2}}{\:\sqrt{\varphi}}\left[\theta\right]_{\mathrm{0}} ^{\lambda_{{max}} } \\ $$$$\lambda=\frac{\mathrm{2}{a}_{\mathrm{1}} }{\varphi}\mathrm{sin}\:^{\mathrm{2}} \theta \\ $$$$\mathrm{sin}\:\theta=\sqrt{\frac{\lambda\varphi}{\mathrm{2}{a}_{\mathrm{1}} }}\Rightarrow\theta=\mathrm{sin}^{−\mathrm{1}} \left(\sqrt{\frac{\lambda\varphi}{\mathrm{2}{a}_{\mathrm{1}} }}\right) \\ $$$$\Rightarrow\frac{\mathrm{2}}{\:\sqrt{\varphi}}\left[\mathrm{sin}^{−\mathrm{1}} \left(\sqrt{\frac{\lambda\varphi}{\mathrm{2}{a}_{\mathrm{1}} }}\right)\right]_{\mathrm{0}} ^{\lambda_{{max}} } \\ $$$${t}=\frac{\mathrm{2}}{\:\sqrt{\left(\xi{g}+\frac{\delta}{{M}_{\mathrm{1}} }\right)}}\mathrm{sin}^{−\mathrm{1}} \left(\sqrt{\frac{\left(\frac{\mathrm{8}{M}_{\mathrm{2}} {M}_{\mathrm{3}} {g}}{\left(\xi{gM}_{\mathrm{1}} +\delta\right)\left({M}_{\mathrm{3}} +{M}_{\mathrm{2}} \right)}\right)\left(\xi{g}+\frac{\delta}{{M}_{\mathrm{1}} }\right)}{\mathrm{2}\frac{\mathrm{4}{M}_{\mathrm{2}} {M}_{\mathrm{3}} {g}}{{M}_{\mathrm{1}} \left({M}_{\mathrm{2}} +{M}_{\mathrm{3}} \right)}}}\right) \\ $$$${M}_{\mathrm{1}} =\mathrm{2}{M}\:,{M}_{\mathrm{2}} ={M},{M}_{\mathrm{3}} =\mathrm{2}{M}\:\mu=\mathrm{0} \\ $$$${t}=\frac{\mathrm{2}}{\:\sqrt{\frac{\delta}{\mathrm{2}{M}}}}\mathrm{sin}^{−\mathrm{1}} \left(\sqrt{\frac{\frac{\mathrm{16}{Mg}}{\mathrm{3}\delta}×\frac{\delta}{\mathrm{2}{M}}}{\mathrm{2}×\frac{\mathrm{4}}{\mathrm{3}}{g}}}\right) \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{2}{M}}}{\:\sqrt{\delta}}\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{1}\right)=\frac{\pi}{\mathrm{2}}\mathrm{2}\sqrt{\frac{\mathrm{2}{M}}{\delta}} \\ $$$${So}\:{t}=\pi\sqrt{\frac{\mathrm{2}{M}}{\delta}}\:{when}\:\lambda=\lambda_{{max}} \\ $$
Commented by fantastic last updated on 03/Nov/25
also going to include t, v_(max) ,x at v_(max) , t at v_(max)
$${also}\:{going}\:{to}\:{include}\:{t},\:{v}_{{max}} \:,{x}\:{at}\:{v}_{{max}} ,\:{t}\:{at}\:{v}_{{max}} \\ $$
Commented by mr W last updated on 03/Nov/25
fundamental serious mistake! if y and x are a function of t respectively, and (dy/dt)=a+bx+cxt you can not say y=at+bxt+(1/2)cxt^2 because x isn′t a constant with respect to t.
$${fundamental}\:{serious}\:{mistake}! \\ $$$${if}\:{y}\:{and}\:{x}\:{are}\:{a}\:{function}\:{of}\:{t}\: \\ $$$${respectively},\:{and} \\ $$$$\frac{{dy}}{{dt}}={a}+{bx}+{cxt} \\ $$$${you}\:{can}\:{not}\:{say} \\ $$$${y}={at}+{bxt}+\frac{\mathrm{1}}{\mathrm{2}}{cxt}^{\mathrm{2}} \\ $$$${because}\:{x}\:{isn}'{t}\:{a}\:{constant}\:{with} \\ $$$${respect}\:{to}\:{t}. \\ $$
Commented by fantastic last updated on 03/Nov/25
ye.You know,sometime mistakes happen :)
$${ye}.{You}\:{know},{sometime} \\ $$$$\left.{mistakes}\:{happen}\::\right) \\ $$
Commented by mr W last updated on 03/Nov/25
(dv/dt)=((4M_2 M_3 g)/(M_1 (M_2 +M_3 )))−ξλg−((δλ)/M_1 ) v≠a_1 t−ξλgt−((δλ)/M_1 )t [after integrating]
$$\frac{{dv}}{{dt}}=\frac{\mathrm{4}{M}_{\mathrm{2}} {M}_{\mathrm{3}} {g}}{{M}_{\mathrm{1}} \left({M}_{\mathrm{2}} +{M}_{\mathrm{3}} \right)}−\xi\lambda{g}−\frac{\delta\lambda}{{M}_{\mathrm{1}} } \\ $$$${v}\neq{a}_{\mathrm{1}} {t}−\xi\lambda{gt}−\frac{\delta\lambda}{{M}_{\mathrm{1}} }{t}\:\left[{after}\:{integrating}\right] \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *