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Intro to Simple Pendulum

A simple pend...

Question

A simple pendulum has a time period of 3.0 s. If the point of suspension of the pendulum starts moving vertically upward with a velocity v = kt where $k = 4.4 m s^{- 2}$ , the new time period will be (Take g = 10 $m s^{- 2}$

$\frac{9}{4} s$

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$\frac{5}{3} s$

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$2.5 s$

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$4.4 s$

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Solution

The correct option is C
$2.5 s$

Given, $v = 4.4 t \Rightarrow a = 4.4 m / s^{2}$
When the point of suspension starts moving upward, effective acceleration, $g^{'} = g + a = 14.4 m / s^{2}$
Now, $\frac{T^{'}}{T} = \frac{2 π \sqrt{\frac{l}{g^{'}}}}{2 π \sqrt{\frac{l}{g}}} = \frac{g}{g^{'}} = \frac{10}{14.4} \Rightarrow T^{'} = \frac{10}{12} \times 3 = 2.5 s$

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Similar questions

A simple pendulum has a time period of 3.0 s. If the point of suspension of the pendulum starts moving vertically upward with a velocity v = kt where $k = 4.4 m s^{- 2}$ , the new time period will be (Take g = 10 $m s^{- 2}$

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\frac{T_{1}^{2}}{T_{2}^{2}} = ? (g = 10 m / s^{2})

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A simple pendulum has a time period of 3.0 s. If the point of suspension of the pendulum starts moving vertically upward with a velocity v = kt where k=4.4ms−2, the new time period will be (Take g = 10 ms−2

The correct option is C 2.5 s Given, v=4.4t⇒a=4.4 m/s2 When the point of suspension starts moving upward, effective acceleration, g′=g+a=14.4 m/s2 Now, T′T=2π√lg′2π√lg=gg′=1014.4⇒T′=1012×3=2.5 s

A simple pendulum has a time period of 3.0 s. If the point of suspension of the pendulum starts moving vertically upward with a velocity v = kt where $k = 4.4 m s^{- 2}$ , the new time period will be (Take g = 10 $m s^{- 2}$