A uniform rod of length 2.0 ,m is suspended through an end and is set into oscillation with small am

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13.Oscillations

medium

A uniform rod of length $2.0 \,m$ is suspended through an end and is set into oscillation with small amplitude under gravity. The time period of oscillation is approximately .... $\sec$

$1.60$

$1.80$

$2$

$2.40$

Solution

(d)This is the special case of physical pendulum and in this case $T = 2\pi \sqrt {\frac{{2l}}{{3g}}} $
$ \Rightarrow $ $T = 2 \times 3.14\sqrt {\frac{{2 \times 2}}{{3 \times 9.8}}} = 2.31\sec \approx 2.4\sec $

Standard 11

Physics

A simple pendulum of length $L$ and mass (bob) $M$ is oscillating in a plane about a vertical line between angular limits $ – \varphi $ and $ + \varphi $. For an angular displacement $\theta (|\theta | < \varphi )$, the tension in the string and the velocity of the bob are $T$ and $ v$ respectively. The following relations hold good under the above conditions

hard

(IIT-1986)

Can simple pendulum clock be used in an artificial satellite ? Why ?

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If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is $\frac{x}{2}$ times its original time period. Then the value of $x$ is:

medium

(NEET-2024)

The period of oscillation of a simple pendulum of constant length at earth surface is $T$. Its period inside a mine is

easy

A simple pendulum has time period 't'. Its time period in a lift which is moving upwards with acceleration $3 ms ^{-2}$ is

easy

A uniform rod of length $2.0 \,m$ is suspended through an end and is set into oscillation with small amplitude under gravity. The time period of oscillation is approximately .... $\sec$

Solution

Similar Questions

Can simple pendulum clock be used in an artificial satellite ? Why ?

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