Tag: oscillations

Questions Related to oscillations

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$\ m/s^2$)

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

  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

Show answer
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

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

  1. 0.5 N

  2. 1 N

  3. 2 N

  4. 4 N

Show answer
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.

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

  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 

Show answer
Correct Option: C
Explanation:

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

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

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

  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.

Show answer
Correct Option: C
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 $X=6m$

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:

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

  1. $-ma\omega^2\cos{\omega t}$

  2. $ma\omega^2\cos{\omega t}$

  3. $ma\omega^2\sin{\omega t}$

  4. $mg+ma\omega^2\cos{\omega t}$

Show answer
Correct Option: D
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 } $