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Complete the following seutence In a damped oscillation of a pendulam .........

A. the sum of potential energy and kinetic energy is conserved. B. mechanical energy is not conserved C. the kinetic energy is conserved D. the potential energy is conserved

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Explanation

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If the time period of undamped oscillation is T and that of damped oscillatin is $T^1$ ,then what is the relation between T &

A. $ T^1 \lt T $ B. nothing can be said, unless the drag force constant is known. C. $T^1 = T $ D. $ T^1 \gt T $

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Explanation

$ T \alpha { 1 \over \omega } \omega = \sqrt { k \over m } \Rightarrow 2 \pi \sqrt { m \over k } $ $ where as T' \alpha { 1 \over \omega '} where \omega' = \sqrt { {k\over m} = { b^2 \over 4m^2} } $ $as \omega' \lt \omega \Rightarrow T' \gt T $

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The energy dissiplated in a damped oscillation ...........

A. decays exponentially B. decay curve will depend on the drag constant C. decays linerly D. decays following a sine curve with diminishing amplitude.

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Explanation

$ E = E_0 e ^ { - \lambda t } $

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What is the equatin for a damped oscillator, where k and b are constants and x is displacement.

A. $ { md^2 x \over dt^2 } = kx + { bdx \over dt } = 0 $ B. $ { md^2 x \over dt^2 } xkx = { bdx \over dt } $ C. $ { md^2 x \over dt^2 } = kx + { bdx \over dt } $ D. $ { md^2 x \over dt^2 } - kx = b { dx \over dt } $

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Explanation

$ F = - kx -by \Rightarrow m { d^2 x \over dt^2 } + { bdx \over dt } + kx =0$

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In a damped oscillatin with damping constant b. The time takenfor amplitude of oscillatin to drop to half what is its initial value ?

A. $ { b \over m } ln 2 $ B. $ { b \over 2m } ln 2 $ C. $ { m \over b } ln 2 $ D. $ { 2m \over b } ln 2 $

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Explanation

$ A = A_0 e ^ {-bt /m} , t = T _{1/2 } whrn A = { A_0 \over 2 } $

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In a damped oscillatin with damping constant b. The time takenfor its mechanical energy to drop to half. What is its value ?

A. $ { b \over m } ln 2 $ B. $ { b \over 2m } ln 2 $ C. $ { m \over b } ln 2 $ D. $ { 2m \over b } ln 2 $

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Explanation

$ E =E_0 e ^ { -bt / m } when E = { E_0 \over 2 } , t = T _ {1/2} $

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For a pendulum in undamped oscillation, with a bob of mass mand radius r, with a string of length l . What is the time period ?

A. $ T = 2 \pi \sqrt { l \over g } $ B. T depends on m C. $ T \gt 2 \pi \sqrt { l \over g } $ D. $ T \lt 2 \pi \sqrt { l \over g } $

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Explanation

for an ideal pendulum , r --> 0 , but for a finite r , $ T = 2 \pi \sqrt { I \over mgl } = 2 \pi \sqrt { 2/5 mr^2 + ml^2 \over mgl } $ $ T = 2 \pi \sqrt { l \over g } where r --> 0 $ otherwise , $ T \gt 2 \pi \sqrt { l \over g } $

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In the experiment of simple pendulum, we have taken a thread of 140 cm, and an amplitude of 5 cm to begin with. Here $ \theta $ to begin with is about............

A. $ 5 ^\circ $ B. $ 8 ^circ $ C. $ 2 ^circ $ D. $ 3 ^circ $

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Explanation

$ \theta = { l \over r } = { 5 cm \over 140 cm } = { 1 \over 28 } rad = 2.05 ^\circ \cong 2^\circ $

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In the experiment we of simple pendulum keep $ \theta \lt 5 ^\circ $ , so as ensure.......

A. mass m does not interfere in the time period B. g remains constant C. the air drag is not too much D. $ sin \theta \cong \theta $ where by motion becomes simple harmonic.

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Explanation

$ IF sin \theta \cong \theta $ $ F = - mg sin \theta = - mg \theta $ $ \Rightarrow F = - ( {mg \over l })x $ $ or F \alpha - x $

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When the moment of force is maximum, then what is the angle between force and position vector of the force ?

A. $ 90 ^\circ $ B. $ 0 ^\circ $ C. $ 30 ^\circ $ D. $ 45 \circ $

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Explanation

$ \tau = Fr sin \theta $ $ ( sin \theta)_ { max} = 1 for \theta = 90 $

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