Question Number 84231 by Rio Michael last updated on 10/Mar/20
![A compound pendulum oscillates though a small angle θ about its equilibrium position such that 10a((dθ/dt))^2 = 4g cos θ , a >0 . its period is [A] 2π(√(((5a)/(4g)) )) [B] ((2π)/5)(√(a/g)) [C] 2π(√(((2g)/(5a)) )) [D] 2π(√((5a)/g))](Q84231.png)
$$\mathrm{A}\:\mathrm{compound}\:\mathrm{pendulum}\:\mathrm{oscillates}\:\mathrm{though}\:\mathrm{a} \\ $$ $$\mathrm{small}\:\mathrm{angle}\:\theta\:\mathrm{about}\:\mathrm{its}\:\mathrm{equilibrium}\:\mathrm{position} \\ $$ $$\mathrm{such}\:\mathrm{that} \\ $$ $$\:\:\:\mathrm{10}{a}\left(\frac{{d}\theta}{{dt}}\right)^{\mathrm{2}} \:=\:\mathrm{4}{g}\:\mathrm{cos}\:\theta\:,\:{a}\:>\mathrm{0}\:.\:\mathrm{its}\:\mathrm{period}\:\mathrm{is}\: \\ $$ $$\left[\mathrm{A}\right]\:\mathrm{2}\pi\sqrt{\frac{\mathrm{5}{a}}{\mathrm{4}{g}}\:\:}\:\:\:\left[\mathrm{B}\right]\:\frac{\mathrm{2}\pi}{\mathrm{5}}\sqrt{\frac{{a}}{{g}}}\:\:\left[\mathrm{C}\right]\:\mathrm{2}\pi\sqrt{\frac{\mathrm{2}{g}}{\mathrm{5}{a}}\:}\:\:\left[\mathrm{D}\right]\:\mathrm{2}\pi\sqrt{\frac{\mathrm{5}{a}}{{g}}}\: \\ $$
Answered by TANMAY PANACEA last updated on 10/Mar/20
$$\frac{{d}\theta}{{dt}}={w}=\frac{\mathrm{2}\pi}{{T}}\rightarrow{T}=\frac{\mathrm{2}\pi}{{w}} \\ $$ $${w}^{\mathrm{2}} =\left(\frac{{d}\theta}{{dt}}\right)^{\mathrm{2}} =\frac{\mathrm{4}{g}}{\mathrm{10}{a}}{cos}\theta\approx\frac{\mathrm{4}{g}}{\mathrm{10}{a}}×\mathrm{1} \\ $$ $${w}=\sqrt{\frac{\mathrm{2}{g}}{\mathrm{5}{a}}}\:\rightarrow{T}=\frac{\mathrm{2}\pi}{\sqrt{\frac{\mathrm{2}{g}}{\mathrm{5}{a}}}} \\ $$ $${T}=\mathrm{2}\pi×\sqrt{\frac{\mathrm{5}{a}}{\mathrm{2}{g}}}\:\:\:\rightarrow\:{i}\:{think}\:{so}\: \\ $$
Commented byRio Michael last updated on 10/Mar/20
$$\mathrm{thanks},{i}'\mathrm{ve}\:\mathrm{gotten}\:\mathrm{the}\:\mathrm{idea} \\ $$
Commented bymr W last updated on 11/Mar/20
![answer is wrong sir! ω here is not the angular velocity, i.e. ω≠(dθ/dt) ω is the angular frequency. we know the angular velocity (dθ/dt) is not constant, otherwise there is no oscillation! but ω is a constant! in fact ω here is defined as (√((elasticity)/(mass))). please refer to your physics book. correct answer: 10a((dθ/dt))^2 =4g cos θ ⇒(dθ/dt)=(√((2g)/(5a)))×(√(cos θ)) ⇒(d^2 θ/dt^2 )=(√((2g)/(5a)))×((−sin θ)/(2(√(cos θ))))×(dθ/dt) ⇒(d^2 θ/dt^2 )=(√((2g)/(5a)))×((−sin θ)/(2(√(cos θ))))×(√((2g)/(5a)))×(√(cos θ)) ⇒(d^2 θ/dt^2 )=−((g sin θ)/(5a))≈−(g/(5a))θ ⇒(d^2 θ/dt^2 )+(g/(5a))θ=0 ⇒(d^2 θ/dt^2 )+ω^2 θ=0 with ω=(√(g/(5a))) period T=((2π)/ω)=2π(√((5a)/g)) ⇒answer [D] is correct.](Q84264.png)
$${answer}\:{is}\:{wrong}\:{sir}! \\ $$ $$\omega\:{here}\:{is}\:{not}\:{the}\:{angular}\:{velocity},\:{i}.{e}. \\ $$ $$\omega\neq\frac{{d}\theta}{{dt}} \\ $$ $$\omega\:{is}\:{the}\:{angular}\:{frequency}. \\ $$ $${we}\:{know}\:{the}\:{angular}\:{velocity}\:\frac{{d}\theta}{{dt}}\:{is} \\ $$ $${not}\:{constant},\:{otherwise}\:{there}\:{is}\:{no} \\ $$ $${oscillation}!\:{but}\:\omega\:{is}\:{a}\:{constant}! \\ $$ $${in}\:{fact}\:\omega\:{here}\:{is}\:{defined}\:{as}\:\sqrt{\frac{{elasticity}}{{mass}}}. \\ $$ $${please}\:{refer}\:{to}\:{your}\:{physics}\:{book}. \\ $$ $$ \\ $$ $${correct}\:{answer}: \\ $$ $$\mathrm{10}{a}\left(\frac{{d}\theta}{{dt}}\right)^{\mathrm{2}} =\mathrm{4}{g}\:\mathrm{cos}\:\theta \\ $$ $$\Rightarrow\frac{{d}\theta}{{dt}}=\sqrt{\frac{\mathrm{2}{g}}{\mathrm{5}{a}}}×\sqrt{\mathrm{cos}\:\theta} \\ $$ $$\Rightarrow\frac{{d}^{\mathrm{2}} \theta}{{dt}^{\mathrm{2}} }=\sqrt{\frac{\mathrm{2}{g}}{\mathrm{5}{a}}}×\frac{−\mathrm{sin}\:\theta}{\mathrm{2}\sqrt{\mathrm{cos}\:\theta}}×\frac{{d}\theta}{{dt}} \\ $$ $$\Rightarrow\frac{{d}^{\mathrm{2}} \theta}{{dt}^{\mathrm{2}} }=\sqrt{\frac{\mathrm{2}{g}}{\mathrm{5}{a}}}×\frac{−\mathrm{sin}\:\theta}{\mathrm{2}\sqrt{\mathrm{cos}\:\theta}}×\sqrt{\frac{\mathrm{2}{g}}{\mathrm{5}{a}}}×\sqrt{\mathrm{cos}\:\theta} \\ $$ $$\Rightarrow\frac{{d}^{\mathrm{2}} \theta}{{dt}^{\mathrm{2}} }=−\frac{{g}\:\mathrm{sin}\:\theta}{\mathrm{5}{a}}\approx−\frac{{g}}{\mathrm{5}{a}}\theta \\ $$ $$\Rightarrow\frac{{d}^{\mathrm{2}} \theta}{{dt}^{\mathrm{2}} }+\frac{{g}}{\mathrm{5}{a}}\theta=\mathrm{0} \\ $$ $$\Rightarrow\frac{{d}^{\mathrm{2}} \theta}{{dt}^{\mathrm{2}} }+\omega^{\mathrm{2}} \theta=\mathrm{0}\:{with}\:\omega=\sqrt{\frac{{g}}{\mathrm{5}{a}}} \\ $$ $${period}\:{T}=\frac{\mathrm{2}\pi}{\omega}=\mathrm{2}\pi\sqrt{\frac{\mathrm{5}{a}}{{g}}} \\ $$ $$\Rightarrow{answer}\:\left[{D}\right]\:{is}\:{correct}. \\ $$
Commented byTANMAY PANACEA last updated on 11/Mar/20
$${yes}\:{sir}\:{you}\:{are}\:{right}...{so}\:{pls}\:{ignore}\:{my}\:{post} \\ $$