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

If the length of a seconds pendulum is increased by $2$% then what is loss and gain in a day?

  1. losses $764 \ s$
  2. losses $924 \ s$
  3. gains $236 \ s$
  4. losses $864 \ s$
  5. gains $346 \ s$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$T _0=2\pi\sqrt{\cfrac{l}{g}}\T^1=2\pi\sqrt{\cfrac{l+l\times2/100}{g}}\ \cfrac{T _0}{T^1}=\cfrac{\sqrt{100}}{\sqrt{102}}\ T^1=\cfrac{\sqrt{102}}{\sqrt{100}}T _0\T^1=1.0099T _0\approx  1.01T _0\Loss=(1.01-1)T _0=0.01T _0$

In one second, it looses $0.01sec$
$\Rightarrow$ Total time loose in one day$=(0.01\times24\times3600)seconds\=864seconds$

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

A simple pendulum with a bob of mass m swings with an angular amplitude of ${ 60 }^{ 0 }$, when its angular displacement is ${ 30 }^{ 0 }$, the tension of string would be 

  1. $3\sqrt { 3 } mg$
  2. $\frac { 1 }{ 2 } mg(2\sqrt { 3 } -1)$
  3. $\frac { 1 }{ 2 } mg(3\sqrt { 3 } +2)$
  4. $\frac { 1 }{ 2 } mg(3-\sqrt { 2 } )$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice physics measurements and experimentation a few applications of linear shm simple pendulum example of simple harmonic motion

The simple pendulum acts as second's pendulum on earth. Its time on a planet, whose mass and diameter are twice that of earth is:

  1. $\sqrt { 2 } s$
  2. $2\sqrt { 2 } s$
  3. $2s$
  4. $\dfrac { 1 }{ \sqrt { 2 } } s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Time period of second's pendulum is two second.
Second's pendulum is that simple pendulum whose time period of vibration is two seconds. The bob of such pendulum while oscillating passes through the mean position after every one second.
Noe,
Time period of simple pendulum is given by
$T=2\pi \sqrt { \left( \dfrac { l }{ g }  \right)  } $
or  $T\propto \dfrac { 1 }{ \sqrt { g }  } $             ......(i)
but  $g=\dfrac { GM }{ { R }^{ 2 } } $      (on earth)
and  ${ g }^{ \prime  }=\dfrac { G\left( 2M \right)  }{ 4{ R }^{ 2 } } $     (on planet)
$=\dfrac { 1 }{ 2 } \dfrac { GM }{ { R }^{ 2 } } =\dfrac { g }{ 2 } $
Equation (i) gives
$\dfrac { { T }^{ \prime  } }{ T } =\dfrac { \sqrt { g }  }{ \sqrt { { g }^{ \prime  } }  } =\sqrt { 2 } $
or  ${ T }^{ \prime  }=\sqrt { 2 } T$
  $=\sqrt { 2 } \times 2              \left( T=2s \right) $
  $=2\sqrt { 2 } s$

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

The length of a second's pendulum at a place where g = 9.8m/s $\displaystyle ^{2}$ is 90.2 cm. State whether true or false.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Time period of pendulum is:

$T =2\pi \sqrt [  ]{ \cfrac { l }{ g }  } $
$l=\cfrac { T^{ 2 }g }{ 4\pi ^{ 2 } } $
$l=\cfrac { 4\times 9.8 }{ 4\times \pi ^{ 2 } } $
$l=0.993m=99.3m$
$l$= length of pendulum 
$g$= $9.8m/s$
$T$ = Time period of seconds pendulum $=2s$
So, our given statement is false.

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

The length of a second pendulum at the surface of earth is $1\ m$. The length of second pendulum at the surface of moon, where $g$ is $\dfrac{1}{6} th$ that of earth's surface.

  1. $\dfrac{1}{6} m$
  2. $6 m$
  3. $\dfrac{1}{36}m$
  4. $36 m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know that the time period of a second pendulum is $2s$ .

At earth ,
                $T=2\pi \sqrt{l _{e}/g _{e}}$
                $2=2\pi\sqrt{l _{e}/g _{e}}$
or             $g _{e}=\pi^{2}l _{e}$  ...............................eq1

At moon ,
                $T=2\pi \sqrt{l _{m}/g _{m}}$
or             $2=2\pi \sqrt{l _{m}/(g _{e}/6)}$  , given   $g _{m}=g _{e}/6$
or             $2=2\pi\sqrt{6l _{m}/\pi^{2}l _{e}}$   ,   putting the value of $g _{m}$ from  eq1
or             $l _{m}=l _{e}/6$
Now , given  $l _{e}=1m$

Hence ,     $l _{m}=1/6m$

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

The length of the simple pendulum which ticks seconds is:

  1. $0.5$m
  2. $1$m
  3. $1.5$m
  4. $2$m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The time period of a simple pendulum is
$T = 2 \pi \sqrt{\dfrac{L}{g}}$
where L is the length of the pendulum.
or $ L = \dfrac{gT^2}{4 \pi^2}$
The time period of the simple pendulum which ticks seconds is $2$s.
$\therefore T = 2s$
Substituting in (i), we get


$L = \dfrac{(9.8 m s^{-2})(2s)^2}{4 \times (3.14)^2} = 1m$

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

A second's pendulum is mounted in a rocket. Its period of oscillation will decrease when the rocket is:

  1. moving up with uniform velocity

  2. moving up with uniform acceleration

  3. moving down with uniform acceleration

  4. moving around the earth in a geostationary orbit
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Correct answer= B

As the rocket accelerates upwards, pseudo force acts in the opposite direction of propagation.
=> Pseudo force acts in downward direction and gets added up to gravitational force.
=> Effective gravity= gravitational force+ pseudo force
                                >Gravitational force
=>             g'        >         g       where g' = effective gravity
Since time period of oscillation of pendulum= √(L/g)
       where L= length of the pendulum
                   g= gravitational force acting on the pendulum
=> In this situation,
          time period of oscillation of pendulum=√(L/g')
Since g' > g
=>     √(L/g')      <     √(L/g)
=>  Time period of oscillation decreases