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Blazing goIITian

Joined:	22 Apr 2007
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19 Jul 2009 14:16:02 IST

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A long uniform rod of mass M and length l is attached at one end to a spring of spring constant

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A long uniform rod of mass M and length l is attached at one end to a spring of spring constant k and the other end is hinged. It is displaced slightly and allowed to oscillate. The time period of oscillations is

(A) 2(pi) root(M/k)

(B) 2(pi) root(M/2k)

(D) 2(pi) root(2M/3k)

(E) None of these.

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Comments (4)

'',' Off to a nu start!!

Blazing goIITian

Joined: 13 Jan 2007

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19 Jul 2009 14:23:17 IST

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use energy method...lower d rod by dx .write expression of energy..change in theta wud be = dx/L..rest procedure i think is same!!

n as far as i remember answer shud be D..2m/3k..although not sure..

adi

Blazing goIITian

Joined: 29 Apr 2009

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19 Jul 2009 17:57:30 IST

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answer is c...torque about hinge is kl(theta).t=((ml^2)/3)*(alpha)

Aakash Anuj

Blazing goIITian

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19 Jul 2009 22:33:07 IST

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Ans is B. Take the torque about the other end . The moment of inertia will become ml^2/3 and u can find alpha in terms of theta which is the required condition for shm

ramyani chakrabarty

Blazing goIITian

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20 Jul 2009 01:10:20 IST

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solved by Karthik M

Assuming the spring is displaced by the angle shown in the figure you provided, we have the following :

Torque provided by the spring (Restoring in nature) = $k(lsin\theta)l = kl^2 sin\theta$

From Newton's second law,

$I\alpha = kl^2 sin\theta$

$\frac{ml^2}{3}\alpha = kl^2 sin\theta$

$\alpha = \frac{3ksin\theta}{m} = \frac{3k\theta}{m}$ (For small oscillations)

Comparing with $\alpha = -\omega ^2 \theta$ , and using $\omega = \frac{2\pi}{T}$

T = $2 \pi \sqrt{\frac{m}{3k}}$

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