Tag: example of simple harmonic motion

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

An object is attached to the bottom of a light vertical spring and set vibrating. The maximum speed of the object is 15 ${ cms }^{ -1 }$ and the period is 628 milli-seconds. The amplitude of the motion in centimeters is :

  1. 3.0

  2. 2.0

  3. 1.5

  4. 1.0
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given,


$T=628ms=0.628s$


$v _{max}=15cm/s=0.15m/s$

The maximum speed of the object is given by

$v _{max}=A\omega=A\dfrac{2\pi}{T}$

Amplitude, $A=\dfrac{v _{max}T}{2\pi}$

$A=\dfrac{0.15\times 0.628}{2\times 3.14}=0.015 m$

$A=1.5cm$

The correct option is C.
Multiple choice physics oscillations a few applications of linear shm simple pendulum example of simple harmonic motion

The different equation for linear SHM of a partial of mass $2g$ is $\dfrac {d^{2}x}{dt^{2}} + 16x = 0$. Find the force constant. $[K = mw^{2}]$.

  1. $0.02\ N/m$.
  2. $0.032\ N/m$.
  3. $0.132\ N/m$.
  4. $0.232\ N/m$.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation is d^2x/dt^2 + 16x = 0. Comparing this to d^2x/dt^2 + w^2x = 0, we get w^2 = 16, so w = 4 rad/s. Given mass m = 2g = 0.002 kg, the force constant K = m * w^2 = 0.002 * 16 = 0.032 N/m.

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

If a body mass $36 gm$ moves with S,H,M of amplitude $A=13$ and period  $T=12 sec$. At a time $t=0$ the displacement is $x=+13 cm$. The shortest time of passage from $x=+6.5$ cm to $x=-6.5$ is

  1. 4 sec

  2. 2 sec

  3. 6 sec

  4. 3 sec
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{array}{l} m=3bg,\, A=13,T=125 \ displacementx\left( t \right) =13\sin  \left( { \frac { { 2\pi t } }{ T }  } \right)  \end{array}$

Shortest time is at maximum slope which crosses zero. It will be from $ - 6.5\,\,to\,\,6.5\,\,$ or 2 times from $0\,to\,\,6.5$
$\begin{array}{l} 6.5=13\sin  \left( { \frac { { 2\pi t } }{ { 12 } }  } \right)  \ 0.5=\sin  \left[ { \left( { \frac { \pi  }{ 6 }  } \right) t } \right]  \ t=1\, \sec   \ total\, \, time=\, 2\times 1=2 \end{array}$

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

A function of time given by $\left(\sin{\omega t}-\cos{\omega t}\right)$ represents

  1. simple harmonic motion

  2. non-periodic motion

  3. periodic but not simple harmonic motion

  4. oscillatory but not simple harmonic motion
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\begin{array}{l} \sin  \omega t-\cos  \omega t \ =\sqrt { 2 } \left[ { \frac { 1 }{ { \sqrt { 2 }  } } \sin  \omega t-\frac { 1 }{ { \sqrt { 2 }  } } \cos  \omega t } \right]  \ =\sqrt { 2 } \left[ { \sin  \omega t\times \cos  \frac { \pi  }{ 4 } -\cos  \omega t\times \sin  \frac { \pi  }{ 4 }  } \right]  \ =\sqrt { 2 } \sin  \left( { \omega t-\frac { \pi  }{ 4 }  } \right)  \ this\, \, function\, \, represents\, \, SHM\, \, as\, \, it\, \, can\, \, be\, \, written\, \, in\, \, the\, \, form: \ a\sin  \left( { \omega t+\phi  } \right)  \ its\, \, period\, \, is,\, \, \frac { { 2\pi  } }{ \omega  }  \end{array}$

Hence,
option $(A)$ is correct answer.

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

A particle is subjected to two simple harmonic motions along $x$ and $y$ directions according to $x=3\sin\ 100\pi t$ $y=4\sin\ 100\pi t$

  1. Motion of particle will be on ellipse travelling in clockwise direction.
  2. Motion of particle will be on a straight line with slope $4/3$
  3. Motion will be simple harmonic motion with amplitude $5$.
  4. Phase difference between two motions is $\pi/2$.
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics oscillatory motion a few applications of linear shm simple pendulum example of simple harmonic motion

A ring whose diameter is 1 meter, oscillates simple harmonically in a vertical plane about a nail fixed at its circumference and perpendicular to plane of ring. The time period will be

  1. 1/4 sec

  2. 1/2 sec

  3. 2sec

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a physical pendulum (ring oscillating about a nail on its circumference), the time period T = 2*pi * sqrt(I / mgd). Here, I = I_cm + md^2 = (1/2)mr^2 + mr^2 = (3/2)mr^2. With d = r, T = 2*pi * sqrt((3/2)mr^2 / mgr) = 2*pi * sqrt(3r / 2g). With diameter 1m, r = 0.5m. T = 2*pi * sqrt(1.5 / 9.8) approx 2 seconds.

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

A person weighing $60\ kg$ stands on a platform which oscillates up and down at a frequency of $2\ Hz$ and amplitude $5\ cm$. The maximum and minimum apparent weights are nearly: ($g$ = 10$\ m/s^2$)

  1. $108$ kg-wt, $12$ kg-wt
  2. $108$ kg-wt, $24$ kg-wt
  3. $54$ kg-wt, $12$ kg-wt
  4. $54$ kg-wt, $24$ kg-wt
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$a=\omega^{2}x$
So, $a _{max}=$ $\omega^{2}A$
We know that $\omega=2\pi  f$
So, $a _{max}=\dfrac{(2\pi\times 2)^{2}\times 5}{100}$
Case I:
$N-mg=ma _{max}$
$N=m(a _{max}+g)$
$=60(10+\dfrac{16\times \pi^{2}\times 5}{100})$
$=1080 $ 

$ N=108$ kg-wt

Case II:
$mg-N=ma _{max}$
or, $N=mg-ma _{max}$
$=60(10-8)$
$=120\ N=12$ kg-wt

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

A body of mass $0.5$ kg is performing S.H.M. with a time period $\pi /2$ seconds. If its velocity at mean position is $1$ m/s, the restoring force acts on the body at a phase angle $60^o$ from extreme position is

  1. 0.5 N

  2. 1 N

  3. 2 N

  4. 4 N
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$T=\dfrac{ \pi}{2}$
$V _{max}=1 m/sec$
$\omega =\dfrac{2\pi}{T}$
$V=4  rad/sec$
$V _{max=}A\omega$
$A\times 4=1$
$A=\dfrac {1}{4}$
$a=\omega^{2}x$
$F= m\omega^{2}x$
$x=A   cos   60^o$
$\therefore x=\dfrac {A}{2}$
$F=0.5\times (4)^{2}\times \dfrac {1}{4}\times \dfrac{1}{2}$
$F=1N$

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

Assertion : If a block is in SHM, and a new constant force acts in the direction of change, the mean position may change.
Reason :In SHM only variable forces should act on the body, for example spring force.

  1. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

  2. Both Assertion and Reason are correct and Reason is not the correct explanation for Assertion

  3. Assertion is correct and Reason is incorrect

  4. Assertion is incorrect and Reason is correct

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In SHM a constant force brings no effective change in the motion but the mean position accelerates.