Tag: physics

Questions Related to physics

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

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

  1. losses $764 \ s$

  2. losses $924 \ s$

  3. gains $236 \ s$

  4. losses $864 \ s$

  5. gains $346 \ s$

Show answer
Correct Option: D
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$=(0.01\times24\times3600)seconds\=864seconds$

If the length of second's pendulum is increased by $2\%$, how many second will it lose per day?

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

  1. $923$ s

  2. $3727$ s

  3. $3642$ s

  4. $864$ s

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Correct Option: D

The different equation of simple harmonic motion for a seconds pendulum is:

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

  1. $\dfrac{d^2 x}{dt^2} + x = 0$

  2. $\dfrac{d^2 x}{dt^2} + \pi x = 0$

  3. $\dfrac{d^2 x}{dt^2} + 4 \pi x = 0$

  4. $\dfrac{d^2 x}{dt^2} + \pi^2 x = 0$

Show answer
Correct Option: A

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 

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

  1. $3\sqrt { 3 } mg$

  2. $\frac { 1 }{ 2 } mg(2\sqrt { 3 } -1)$

  3. $\frac { 1 }{ 2 } mg(3\sqrt { 3 } +2)$

  4. $\frac { 1 }{ 2 } mg(3-\sqrt { 2 } )$

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Correct Option: B

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:

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

  1. $\sqrt { 2 } s$

  2. $2\sqrt { 2 } s$

  3. $2s$

  4. $\dfrac { 1 }{ \sqrt { 2 } } s$

Show answer
Correct Option: B
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) $
  $=2\sqrt { 2 } s$

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.

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

  1. True

  2. False

Show answer
Correct Option: B
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.

A second's pendulum can be used as a timing device

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

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

The second's pendulum has a definite time period of 2 seconds. We can calculate the time by counting its no.of oscillation. So, it can be used as a time measuring equipment.

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.

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

  1. $\dfrac{1}{6} m$

  2. $6 m$

  3. $\dfrac{1}{36}m$

  4. $36 m$

Show answer
Correct Option: A
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 ,     $l _{m}=1/6m$

The length of the simple pendulum which ticks seconds is:

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

  1. $0.5$m

  2. $1$m

  3. $1.5$m

  4. $2$m

Show answer
Correct Option: B
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


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

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

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

  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

Show answer
Correct Option: B
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