Tag: example of simple harmonic motion

Questions Related to 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.

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

  1. II, III and IV

  2. I, II and IV

  3. I, II and III

  4. I, II, III and IV

Show answer
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?

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

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

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

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

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

Show answer
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:

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

  1. $400 Hz$

  2. $100 Hz$

  3. $256 Hz$

  4. $200 Hz$

Show answer
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

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 $X=6m$

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):

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

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

  2. 2

  3. $5\sqrt {10} $

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

Show answer
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:

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 } $

The frequency of a seconds pendulum is equal to :

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

  1. 0.5 Hz

  2. 1 Hz

  3. 2 Hz

  4. 0.1 Hz

Show answer
Correct Option: A
Explanation:

The frequency of seconds pendulum is $= \dfrac{1}{2} = 0.5$ HZ

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

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

  1. $20s$

  2. $50s$

  3. $40s$

  4. $5s$

Show answer
Correct Option: C
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.

The length of a second's pendulum on the surface of the earth is equal to 99.49 cm. True or false.

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

  1. True

  2. False

Show answer
Correct Option: A
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,

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

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

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

  1. 4πG/3gR

  2. 3πR/4gG

  3. 3g/4πRG     

  4. πRg/12G

Show answer
Correct Option: C
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 $g=\cfrac{GM}{R^2}\P=\cfrac{3g}{4\pi RG}$