Tag: a few applications of linear shm

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

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?

  1. $10.3$ sec/day
  2. $19$ sec/day
  3. $5.5$ sec/day
  4. $6.8$ sec/day
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The fractional change in time period is given by delta T / T = 1/2 * alpha * delta theta. With alpha = 12*10^-6 and delta theta = 20 degrees, the fractional change is 1.2*10^-4. Multiplying by the number of seconds in a day (86400), we get approximately 10.36 seconds.

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

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

  1. $99.4\ cm$
  2. $89.4\ cm$
  3. $79.4\ cm$
  4. $109.4\ cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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


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

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

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

  1. $0.25\, m$
  2. $0.50\, m$
  3. $0.15\, m$
  4. $0.30\, m$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.)

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

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

  1. $0.25\ m$
  2. $0.50\ m$
  3. $0.25\ cm$
  4. $0.25\ mm$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.)

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

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? 

  1. 10 vibrations

  2. 11 vibrations

  3. 21 vibrations

  4. 20 vibrations
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

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.

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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. 

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

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?

  1. $1 : 1$
  2. $1 : 2$
  3. $2: 1$
  4. $4 : 3$
Reveal answer Fill a bubble to check yourself
A Correct answer
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$.$