A long stiff uniform wire of length ell is su | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) For a physical pendulum,the time period is given by $T = 2 \pi \sqrt{\frac{I}{mgd}}$,where $I$ is the moment of inertia about the pivot,$m$ is the

Vedclass · Accepted Answer

$\sqrt{\frac{3}{2 \pi}} T$

Vedclass · Answer

(D) For a physical pendulum,the time period is given by $T = 2 \pi \sqrt{\frac{I}{mgd}}$,where $I$ is the moment of inertia about the pivot,$m$ is the mass,$g$ is acceleration due to gravity,and $d$ is the distance from the pivot to the center of mass $(COM)$.For the rod of length $\ell$ suspended from one end:$I = \frac{m\ell^2}{3}$ and $d = \frac{\ell}{2}$.$T = 2 \pi \sqrt{\frac{m\ell^2/3}{mg\ell/2}} = 2 \pi \sqrt{\frac{2\ell}{3g}} \quad \dots(i)$When the wire is bent into a circle of radius $R$,the circumference is $\ell = 2 \pi R$,so $R = \frac{\ell}{2 \pi}$.For a ring suspended from a point on its circumference,the moment of inertia about the pivot is $I' = I_{cm} + mR^2 = mR^2 + mR^2 = 2mR^2$ (using parallel axis theorem).The distance from the pivot to the $COM$ is $d' = R$.$T' = 2 \pi \sqrt{\frac{2mR^2}{mgR}} = 2 \pi \sqrt{\frac{2R}{g}} \quad \dots(ii)$Substituting $R = \frac{\ell}{2 \pi}$ into $(ii)$:$T' = 2 \pi \sqrt{\frac{2}{g} \cdot \frac{\ell}{2 \pi}} = 2 \pi \sqrt{\frac{\ell}{\pi g}}$.Dividing $(ii)$ by $(i)$:$\frac{T'}{T} = \frac{2 \pi \sqrt{2R/g}}{2 \pi \sqrt{2\ell/3g}} = \sqrt{\frac{2R}{g} \cdot \frac{3g}{2\ell}} = \sqrt{\frac{3R}{\ell}} = \sqrt{\frac{3(\ell/2\pi)}{\ell}} = \sqrt{\frac{3}{2 \pi}}$.Thus,$T' = \sqrt{\frac{3}{2 \pi}} T$.

$A$ long stiff uniform wire of length $\ell$ is suspended from one end. The time period of oscillations of the wire is $T$. If the wire is now bent into a circle and suspended from a knife edge so that it can oscillate freely in the plane of the ring,its time period will be

Similar Questions

$A$ body performs $S.H.M.$ under the action of force $F_1$ with period $T_1$ seconds. If the force is changed to $F_2$,it performs $S.H.M.$ with period $T_2$ seconds. If both forces $F_1$ and $F_2$ act simultaneously in the same direction on the body,the period in seconds will be

Under the influence of force $F_1$,a body oscillates with a period $T_1$,and due to another force $F_2$,the body oscillates with a period $T_2$. If both forces act simultaneously,what is the resultant period? (Consider the displacement is the same in all three cases.)

$A$ $U$-tube of a uniform bore has its arms vertical. The total length of the liquid in the two arms of the $U$-tube is $L$. The time period $T$ of the oscillation of the liquid column when it is displaced by $y$ is ($g$ = acceleration due to gravity).

Vedclass Test Series

Exam Paper Generator

Online Exam Module