Tag: a few applications of linear shm

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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.

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

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
Reveal answer Fill a bubble to check yourself
D Correct answer
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.
$f \propto \cfrac{1}{\text{thickness of wire}}$

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

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. $\phi _A > \phi _B > 0 > \phi _c$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

A stretched string of one meter length, fixed at both the ends having mass of $5 \times 10^{-4}$ kg is under tension of 20 N. It is plucked at a point situated 25 cm from one end. The stretched string would vibrate with the frequency of:

  1. $400 Hz$
  2. $100 Hz$
  3. $256 Hz$
  4. $200 Hz$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The fundamental frequency of a stretched string is given by f = (1/2L) * sqrt(T/mu). Given L = 1 m, mass m = 5 x 10^-4 kg so linear mass density mu = 5 x 10^-4 kg/m, and tension T = 20 N. Plugging these in yields f = (1/2) * sqrt(20 / (5 x 10^-4)) = (1/2) * sqrt(40000) = (1/2) * 200 = 100 Hz.

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 } $