Tag: simple pendulum

Questions Related to simple pendulum

The frequency of a second's pendulum is

physics simple harmonic motion a few applications of linear shm simple pendulum example of simple harmonic motion

  1. 0.5 Hz

  2. 1.0 Hz

  3. 1.5 Hz

  4. none of these

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Correct Option: A
Explanation:

Seconds pendulum takes 2 $sec$ to make one complete oscillation. 

Hence, frequency = $\dfrac{1}{2}$ Hz

The frequency of a second's pendulum is :

physics simple harmonic motion a few applications of linear shm simple pendulum example of simple harmonic motion

  1. $0.5\;Hz$

  2. $1.0\;Hz$

  3. $2.0\;Hz$

  4. $5.5\;Hz$

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Correct Option: A
Explanation:

A seconds pendulum is a pendulum whose period is precisely two seconds

A simple pendulum with length $L$ and mass $m$ of the bob is oscillating with an amplitude $a$. 

Then the maximum tension in the string is :

physics simple harmonic motion a few applications of linear shm simple pendulum example of simple harmonic motion

  1. $mg$

  2. $mg[1+(\dfrac{a}{L})^{2}]$

  3. $mg[1+\dfrac{a}{2L}]^{2}$

  4. $mg[1+(\dfrac{a}{L})]^{2}$

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Correct Option: D

There is a clock which gives correct time at $20^o$C is subjected to $40^o$C. The coefficient of linear expansion of the pendulum is $12\times 10^{-6}$ per $^oC$, how much is gain or loss in time?

physics simple harmonic motion a few applications of linear shm simple pendulum example of simple harmonic motion

  1. $10.3$ sec/day

  2. $19$ sec/day

  3. $5.5$ sec/day

  4. $6.8$ sec/day

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Correct Option: A

Find the length of a simple pendulum such that its time period is $2\ s$.

physics simple harmonic motion a few applications of linear shm simple pendulum example of simple harmonic motion

  1. $99.4\ cm$

  2. $89.4\ cm$

  3. $79.4\ cm$

  4. $109.4\ cm$

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Correct Option: A
Explanation:

$T\, =\, 2 \pi\,\sqrt{\displaystyle \frac{L}{g}}\, \Rightarrow\, T^2\, =\, 4 \pi^2\, \times\, \displaystyle \frac{L}{g}$


$T^2\, =\, 4\pi ^2\, \displaystyle \frac {L}{g}$
$\Rightarrow\, 2^2\, =\, 4\, \times\, 3.14\, \times\, 3.14\, \times\, \displaystyle \frac {L}{9.8}$
$\Rightarrow\, L\, =\, \displaystyle \frac {4\, \times\, 9.8}{4\, \times\, 3.14\, \times\, 3.14}\, m\, =\, 0.994\, m\, =\, 99.4\, cm$

A desktop toy pendulum swings back and forth once every $1.0 s$. How long is this pendulum?

physics simple harmonic motion a few applications of linear shm simple pendulum example of simple harmonic motion

  1. $0.25\, m$

  2. $0.50\, m$

  3. $0.15\, m$

  4. $0.30\, m$

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Correct Option: A
Explanation:

$T\, =\, 2 \pi\,\sqrt{\displaystyle \frac{L}{g}}\, \Rightarrow\, T^2\, =\, 4 \pi^2\, \times\, \displaystyle \frac{L}{g}$ .. (1)


Putting $T = 1$ in eqn. (1), 


We get $1\, =\, 4 \pi^2\, \times\, \displaystyle \frac{L}{g}$ $\Rightarrow\, L\, =\, \displaystyle \frac{g}{4\, \pi^2}\, =\, \displaystyle \frac{9.8}{4\, \times\, 3.14\, \times\, 3.14}m\, \Rightarrow\, L\, =\, \displaystyle \frac{9.8}{39.44}m\, =\, 0.2484\, m\, =\, 0.25\, m$ (approx.)

You are designing a pendulum clock to have a period of $1.0\ s$. How long should the pendulum be ?

physics simple harmonic motion a few applications of linear shm simple pendulum example of simple harmonic motion

  1. $0.25\ m$

  2. $0.50\ m$

  3. $0.25\ cm$

  4. $0.25\ mm$

Show answer
Correct Option: A
Explanation:

$T\, =\, 2 \pi\,\sqrt{\displaystyle \frac{L}{g}}\, \Rightarrow\, T^2\, =\, 4 \pi^2\, \times\, \displaystyle \frac{L}{g}$ .. (1)

Putting $T = 1$ in eqn. (1), 

We get $1\, =\, 4 \pi^2\, \times\, \displaystyle \frac{L}{g}$ $\Rightarrow\, L\, =\, \displaystyle \frac{g}{4\, \pi^2}\, =\, \displaystyle \frac{9.8}{4\, \times\, 3.14\, \times\, 3.14}m\, \Rightarrow\, L\, =\, \displaystyle \frac{9.8}{39.44}m\, =\, 0.2484\, m\, =\, 0.25\, m$ (approx.)

Two pendulums of lengths 121 cm and 100 cm start vibrating at the same instant. They are in the mean position and in the same phase. After how many vibrations of the shorter pendulum, the two will be in the same phase in the mean position? 

physics simple harmonic motion a few applications of linear shm simple pendulum example of simple harmonic motion

  1. 10 vibrations

  2. 11 vibrations

  3. 21 vibrations

  4. 20 vibrations

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Correct Option: A
Explanation:

Two pendulums of length $121cm$ and $100cm$.


Let,
$L _1=121cm=\dfrac{121}{100}=1.21m$

$L _2=100cm=\dfrac{100}{100}=1m$

We have to find the vibrations made by the shorter pendulum, such that both will be in same phase from the reaction,

$T _1=longer\,pendulum$


$T _2=shorter\,pendulum$


$T=2\pi\sqrt{\dfrac{L}{g}}$

$T\propto \sqrt{L}$

$\dfrac{T _1}{T _2} \propto \sqrt{{L _1}{L _2}}$

$\dfrac{T _1}{T _2}\propto \sqrt{\dfrac{1.21}{1}}$

$\dfrac{T _1}{T _2}=\dfrac{1.1}{1}$

$10T _1=11T _2$

$10$ vibrations of longer pendulum= $11$ vibrations of shorter pendulum

Assertion (A): A wooden cube of side a floats in a non viscous liquid of density r. When it is slightly pressed and released, then it executes SHM
Reason (R): The net force responsible for SHM is the resultant of buoyancy force and true weight of the body.

physics simple harmonic motion a few applications of linear shm simple pendulum example of simple harmonic motion

  1. Both A and R are true and R is the correct explanation of A

  2. Both A and R are true and R is not the correct explanation of A

  3. A is true and R is false

  4. A is false and R is true

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Correct Option: A
Explanation:

When a wooden cube of side a floats in a non viscous liquid of density r and if it is slightly pressed and released it executes SHM because of the buoyancy force which is acting on the cube in upward direction. And hence the answer. 

A body is broken into two parts of masses $m _1$ and $m _2$ These parts are then separated by a distance r ,What is the value of $m _1/m _2$ so that the gravitational force has maximum possible value?

physics simple harmonic motion a few applications of linear shm simple pendulum example of simple harmonic motion

  1. $1 : 1$

  2. $1 : 2$

  3. $2: 1$

  4. $4 : 3$

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Correct Option: A
Explanation:

Let the mass of the body is $m$

$\therefore \,{m _1} + {m _2} = m$
${F _G} = \frac{{G{m _1}{m _2}}}{{{r^2}}}$
${F _G} = \frac{{G{m _1}\left( {m - {m _1}} \right)}}{{{r^2}}}$
for ${F _G} \to \max \,\frac{{d\left( {{F _G}} \right)}}{{d{m _1}}} = 0$
$ = \frac{G}{{{r^2}}}\left( {m - 2{m _1}} \right) = 0$
$ = {m _1} = \frac{m}{2}$
${m _2} = \frac{m}{2}$
$\therefore {m _1}/{m _2} = 1:1$
Hence$,$ optin $(A)$ is correct$.$