Tag: a few applications of linear shm

Questions Related to a few applications of linear shm

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.

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. $200 Hz$


Correct Option: D

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.


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 $X=6m$

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. $\frac{1}{{5\sqrt {10} }}$


Correct Option: A

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


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 } $