Tag: physics

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

Speed v of a particle moving along a straight line, when it is at a distance x from a fixed point on the line is given by $V^2=108-9x^2$(all quantities in S. I. unit). Then

  1. The motion is uniformly accelerated along the straight line
  2. The magnitude of the acceleration at a distance 3 cm from the fixed point is $0.27m/s^2$
  3. The motion is simple harmonic about $x=6$m

  4. The maximum displacement from fixed point is 4cm.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$V^2=108-9x^2$


for SHM

$V^2=\omega^2(A^2-X^2)$

$V^2=9(12-X^2)$

$W=3,A=2\sqrt{3}$

$v\dfrac{dv}{dx}=9(12-2X)$

$\dfrac{dV}{dX}=0$ at $X=6$

So, it will perform SHM about $X=6m$

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

The equation of motion of a particle of mass $1$ g is $\frac{{{d^2}x}}{{d{t^2}}} + {\pi ^2}x = 0$ where $x$ is displacement (in m) from mean position. The frequency of oscillation is ( in Hz):

  1. $\frac{1}{2}$

  2. 2
  3. $5\sqrt {10} $
  4. $\frac{1}{{5\sqrt {10} }}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation is d^2x/dt^2 + pi^2 * x = 0. Comparing to d^2x/dt^2 + w^2 * x = 0, we get w^2 = pi^2, so w = pi. Since w = 2 * pi * f, then pi = 2 * pi * f, which gives f = 1/2 Hz.

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

A planck with a body of mass m placed on to it starts moving straight up with the law $y=a(1-\cos{\omega t})$ where $\omega$ is displacement. Find the time dependent force:

  1. $-ma\omega^2\cos{\omega t}$
  2. $ma\omega^2\cos{\omega t}$
  3. $ma\omega^2\sin{\omega t}$
  4. $mg+ma\omega^2\cos{\omega t}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Total force on the particle will be
$F=mg+m\dfrac { d^{ 2 }y }{ dt^{ 2 } } $
since $y=a(1-\cos { \omega t } )\\ \Rightarrow \dfrac { dy }{ dt } =a\omega \sin { \omega t } \\ \Rightarrow \dfrac { d^{ 2 }y }{ dt^{ 2 } } =a\omega ^{ 2 }\cos { \omega t } $
$\Rightarrow F=mg+ma\omega ^{ 2 }\cos { \omega t } $

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

The time taken to complete $20$ oscillations by a seconds pendulum is: 

  1. $20s$
  2. $50s$
  3. $40s$
  4. $5s$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know that the time period of a seconds pendulum is $T=2$ sec. One second for a swing in one direction and one second for the return swing. 

Thus, time taken to complete one oscillation is $2$ sec.
Hence, time taken to complete 20 oscillations is $2\times 20=40$ sec.

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 on the surface of the earth is equal to 99.49 cm. True or false.

  1. True

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

The time period of seconds pendulum T = 2 seconds, acceleration due to gravity at earth g= 980 $\dfrac { cm }{ { s }^{ 2 } } $,It '$l$' is the length of pendulum,

$l=\dfrac { { T }^{ 2 }g }{ 4{ \pi  }^{ 2 } } \ \Rightarrow l=\dfrac { 4\times 980 }{ 4\times \left( \dfrac { 22 }{ 7 }  \right) ^{ 2 } } =\dfrac { 4\times 980\times 49 }{ 4\times 489 } =99.49$

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

If R is the radius of the earth and g the acceleration due to gravity on the earth's surface, the mean density of the earth is

  1. 4πG/3gR

  2. 3πR/4gG

  3. 3g/4πRG

  4. πRg/12G
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know that

$g=\cfrac{GM}{R^2}$
Also, density $=mass\times volume$
$M=density\times volume\M=P\times\cfrac{4\pi R^3}{3R^2}=P\times\cfrac{4\pi R}{3}$
Put value of m in $g=\cfrac{GM}{R^2}\P=\cfrac{3g}{4\pi RG}$

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

The mass of a bob, suspended in a simple pendulum, is halved from the initial mass, its time period will :

  1. Be less

  2. Be more

  3. Remain unchanged

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

The time period of simple pendulum id given by

$T=2\pi \sqrt{\dfrac{l}{g}}$
where, $l=$ length of simple pendulum
$g=$ acceleration due to gravity
$T=$ Time period
The time period of simple pendulum is independent of the mass of bob, the time period remains unchanged,when mass of bob will change.
The correct option is C.