Tag: physics

Questions Related to physics

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


Correct Option: B
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 $0\,to\,\,6.5$
$\begin{array}{l} 6.5=13\sin  \left( { \frac { { 2\pi t } }{ { 12 } }  } \right)  \ 0.5=\sin  \left[ { \left( { \frac { \pi  }{ 6 }  } \right) t } \right]  \ t=1\, \sec   \ total\, \, time=\, 2\times 1=2 \end{array}$

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


Correct Option: A
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.

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 $\pi/2$.


Correct Option: A

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


Correct Option: C

The graph between restoring force and time in case of SHM is a

  1. parabola

  2. sine curve

  3. straight line

  4. circle


Correct Option: B
Explanation:

We know that for SHM, $x=A\sin(\omega t + \theta)$ and $F=kx=kA\sin(\omega t + \theta)$
Thus it's a sine curve.

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


Correct Option: A
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

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


Correct Option: B
Explanation:

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

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 


Correct Option: C
Explanation:

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

Sitar maestro Ravi Shankar is playing sitar on its strings, and you, as a physicist (unfortunately without musical ears!), observed the following oddities.
I. The greater the length of a vibrating string, the smaller its frequency.
II. The greater the tension in the string, the greater is the frequency.
III. The heavier the mass of the string, the smaller the frequency.
IV. The thinner the wire, the higher its frequency.
The maestro signalled the following combination as correct one.

  1. II, III and IV

  2. I, II and IV

  3. I, II and III

  4. I, II, III and IV


Correct Option: D
Explanation:

Guitar string is a standing wave as both the ends of guitar string are fixed.

$f \propto \cfrac{1}{length}$
$f \propto \sqrt{tension}$
$f \propto \cfrac{1}{\sqrt{\mu}}$, where $\mu$ is mass per unit length.
$f \propto \cfrac{1}{\text{thickness of wire}}$

Three similar oscillators, A, B, C have the same small damping constant $r$, but different natural frequencies $\omega _0 = (k/m)^{\frac{1}{2}} : 1200 Hz, 1800 Hz, 2400 Hz$. If all three are driven by the same source at $1800 Hz$, which statement is correct for the phases of the velocities of the three?

  1. $\phi _A = \phi _B = \phi _c$

  2. $\phi _A < \phi _B = 0 < \phi _c$

  3. $\phi _A > \phi _B = 0 > \phi _c$

  4. $\phi _A > \phi _B > 0 > \phi _c$


Correct Option: A
Explanation:

$tan \phi = \dfrac{\omega L- \dfrac{1}{\omega C}}{R}$

If $\omega < \omega _0 \Longrightarrow$ circuit is capacitive

$\Longrightarrow$ it leads voltage
$\Longrightarrow$ velocity leads force.
if $\omega = \omega _0 \phi = 0$
if $\omega > {\omega} _0$ velocity lags behind force.