Question Number 227789 by mr W last updated on 18/Feb/26
Commented by mr W last updated on 18/Feb/26
$${find}\:{the}\:{period}\:{of}\:{oscillation}. \\ $$
Answered by mahdipoor last updated on 18/Feb/26
![for left mass (disp : u) for right mass (disp : v) mu^(..) =−k(u−v) mv^(..) =k(u−v) get x= [(u),(v) ] ⇒ [(m,0),(0,m) ]x^(..) + [(k,(−k)),((−k),k) ]x= [(0),(0) ] get x=e^(λt) [(a),(b) ]=Ae^(λt) [(1),(T) ] ⇒Ae^(λt) [((λ^2 m+k),(−k)),((−k),(λ^2 m+k)) ] [(1),(T) ]= [(0),(0) ] (char eq) ⇒(λ^2 m+k)^2 −(−k)^2 =0⇒λ^2 =0,−((2k)/m) case 1: λ^2 =0 ⇒ λ=0 (rigid body mod) x=A [(1),(T) ]⇒ T=1 from char eq x_1 =(A+Bt) [(1),(1) ] case 2: λ^2 =−((2k)/m) ⇒ λ=±j(√(2k/m))=±jω (ntural mod) x=Ae^(λt) [(1),(T) ]⇒ T=−1 x_2 =(Ccosω+Dsinω) [(1),((−1)) ] final ans: x=x_1 +x_2](Q227803.png)
$$\mathrm{for}\:\mathrm{left}\:\mathrm{mass}\:\left(\mathrm{disp}\::\:\mathrm{u}\right) \\ $$$$\mathrm{for}\:\mathrm{right}\:\mathrm{mass}\:\left(\mathrm{disp}\::\:\mathrm{v}\right) \\ $$$$\mathrm{m}\overset{..} {\mathrm{u}}=−\mathrm{k}\left(\mathrm{u}−\mathrm{v}\right) \\ $$$$\mathrm{m}\overset{..} {\mathrm{v}}=\mathrm{k}\left(\mathrm{u}−\mathrm{v}\right) \\ $$$$\mathrm{get}\:\:\:\mathrm{x}=\begin{bmatrix}{\mathrm{u}}\\{\mathrm{v}}\end{bmatrix}\:\:\Rightarrow \\ $$$$\begin{bmatrix}{\mathrm{m}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{m}}\end{bmatrix}\overset{..} {\mathrm{x}}+\begin{bmatrix}{\mathrm{k}}&{−\mathrm{k}}\\{−\mathrm{k}}&{\mathrm{k}}\end{bmatrix}\mathrm{x}=\begin{bmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{bmatrix} \\ $$$$\mathrm{get}\:\mathrm{x}=\mathrm{e}^{\lambda\mathrm{t}} \begin{bmatrix}{\mathrm{a}}\\{\mathrm{b}}\end{bmatrix}=\mathrm{Ae}^{\lambda\mathrm{t}} \begin{bmatrix}{\mathrm{1}}\\{\mathrm{T}}\end{bmatrix} \\ $$$$\Rightarrow\mathrm{Ae}^{\lambda\mathrm{t}} \begin{bmatrix}{\lambda^{\mathrm{2}} \mathrm{m}+\mathrm{k}}&{−\mathrm{k}}\\{−\mathrm{k}}&{\lambda^{\mathrm{2}} \mathrm{m}+\mathrm{k}}\end{bmatrix}\begin{bmatrix}{\mathrm{1}}\\{\mathrm{T}}\end{bmatrix}=\begin{bmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{bmatrix}\:\:\:\left(\mathrm{char}\:\mathrm{eq}\right) \\ $$$$\Rightarrow\left(\lambda^{\mathrm{2}} \mathrm{m}+\mathrm{k}\right)^{\mathrm{2}} −\left(−\mathrm{k}\right)^{\mathrm{2}} =\mathrm{0}\Rightarrow\lambda^{\mathrm{2}} =\mathrm{0},−\frac{\mathrm{2k}}{\mathrm{m}} \\ $$$$\mathrm{case}\:\mathrm{1}:\:\lambda^{\mathrm{2}} =\mathrm{0}\:\Rightarrow\:\lambda=\mathrm{0}\:\left(\mathrm{rigid}\:\mathrm{body}\:\mathrm{mod}\right) \\ $$$$\mathrm{x}=\mathrm{A}\begin{bmatrix}{\mathrm{1}}\\{\mathrm{T}}\end{bmatrix}\Rightarrow\:\mathrm{T}=\mathrm{1}\:\mathrm{from}\:\mathrm{char}\:\mathrm{eq} \\ $$$$\mathrm{x}_{\mathrm{1}} =\left(\mathrm{A}+\mathrm{Bt}\right)\begin{bmatrix}{\mathrm{1}}\\{\mathrm{1}}\end{bmatrix}\: \\ $$$$\mathrm{case}\:\mathrm{2}:\:\lambda^{\mathrm{2}} =−\frac{\mathrm{2k}}{\mathrm{m}}\:\Rightarrow\:\lambda=\pm\mathrm{j}\sqrt{\mathrm{2k}/\mathrm{m}}=\pm\mathrm{j}\omega\:\left(\mathrm{ntural}\:\mathrm{mod}\right) \\ $$$$\mathrm{x}=\mathrm{Ae}^{\lambda\mathrm{t}} \begin{bmatrix}{\mathrm{1}}\\{\mathrm{T}}\end{bmatrix}\Rightarrow\:\mathrm{T}=−\mathrm{1} \\ $$$$\mathrm{x}_{\mathrm{2}} =\left(\mathrm{Ccos}\omega+\mathrm{Dsin}\omega\right)\begin{bmatrix}{\mathrm{1}}\\{−\mathrm{1}}\end{bmatrix} \\ $$$$\mathrm{final}\:\mathrm{ans}:\:\mathrm{x}=\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} \\ $$
Commented by mr W last updated on 19/Feb/26
$${thanks}! \\ $$
Answered by mr W last updated on 19/Feb/26
Commented by mr W last updated on 19/Feb/26
$${say}\:{the}\:{length}\:{of}\:{the}\:{massless}\: \\ $$$${spring}\:{is}\:{l}. \\ $$$${since}\:{no}\:{external}\:{force}\:{is}\:{acting} \\ $$$${in}\:{horixontal}\:{direction},\:{the}\:{center} \\ $$$${of}\:{mass}\:{from}\:{m}\:{and}\:{M}\:{remains} \\ $$$${unchanged}.\:{this}\:{point}\:{divides}\:{the} \\ $$$${spring}\:{into}\:{two}\:{parts}\:{with}\:{lengthes} \\ $$$${l}_{\mathrm{1}} \:{and}\:{l}_{\mathrm{2}} . \\ $$$${l}_{\mathrm{1}} =\frac{{ml}}{{M}+{m}},\:{l}_{\mathrm{2}} =\frac{{Ml}}{{M}+{m}} \\ $$$${the}\:{spring}\:{constant}\:{of}\:{the}\:{part}\:{with} \\ $$$${length}\:{l}_{\mathrm{1}} \:{is} \\ $$$${k}_{\mathrm{1}} =\frac{{l}}{{l}_{\mathrm{1}} }×{k}=\frac{\left({M}+{m}\right){k}}{{m}} \\ $$$${the}\:{period}\:{of}\:{oscillation}\:{from}\:{mass} \\ $$$${M}\:{and}\:{spring}\:{k}_{\mathrm{1}} \:{is} \\ $$$${T}=\mathrm{2}\pi\sqrt{\frac{{M}}{{k}_{\mathrm{1}} }}=\mathrm{2}\pi\sqrt{\frac{{Mm}}{\left({M}+{m}\right){k}}}\:\:\checkmark \\ $$