Tag: simple harmonic motion

Questions Related to simple harmonic motion

A force acts on a $30$gm particle in such a way that the position of the particle as a function of time is given by $x=3t-4t^2+t^3$, where x is in metres and t is in seconds. The work done during the first $4$ second is?

physics simple harmonic motion representing shm with circular motion shm as projection of circular motion simple harmonic motion (shm) as a projection of uniform circular motion

  1. $2.88$J

  2. $450$mJ

  3. $490$mJ

  4. $530$mJ

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Correct Option: A
Explanation:

$M=20gm=0.020kg$

$x=3t-4t²+t³m$
$v=dx/dt=3-8t+3t²m/s$
$a=dv/dt=-8+6tm/s²$
$Force=ma=0.120t-0.160N$
$Workdone=dW=Fdx=F\ast \frac { dx }{ dt } \ast dt=(0.120t-0.160)\ast (3-8t+3t²)dt=0.360t³-1.440t²+1.640t-0.480$
$W=0.090t⁴-0.490t³+0.82t²-0.480t=2.88J$

A wheel of radius $1$ meters rolls forward half a revolution on a horizontal ground. The magnitude of the displacement of the point of the wheel initially in contact with the ground is:

physics simple harmonic motion representing shm with circular motion shm as projection of circular motion simple harmonic motion (shm) as a projection of uniform circular motion

  1. $2 \pi$

  2. $\sqrt 2 \pi$

  3. $\sqrt {{\pi ^2} + 4} $

  4. $\pi$

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Correct Option: C
Explanation:
Let fro $A$ to $B$, where complete half a rotation.
we have,
$PQ=\alpha =$ half circumference $=\pi \ m$
and $P' Q=$ diameter $=2\ m$
so we have displacement of $P$
$x=|PP;|=\sqrt {PQ^2 +P; Q^2}$ (Pythagoras theorem)
$\Rightarrow \ x=\sqrt {\pi^2 +4}m$

Two SHMs are represented by the equations 
$y1=10sin(3\Omega t+\frac{\Omega }{4})$ and
$y2=5[sin3\Omega t+\sqrt{3}cos 3\Omega t]$. their amplitudes and in the ratio

physics simple harmonic motion representing shm with circular motion shm as projection of circular motion simple harmonic motion (shm) as a projection of uniform circular motion

  1. 1:2

  2. 2:1

  3. 1:3

  4. 1:1

Show answer
Correct Option: B
Explanation:

Compare the given equations with,

$y=a\sin \omega t$

Where a is amplitude of wave

Here, a1 = 10

 a2 = 5

So the ratio is

$\dfrac{{{a} _{1}}}{{{a} _{2}}}=\dfrac{10}{5}=\dfrac{2}{1}$

Or $2:1$

A simple harmonic oscillator starts from extreme position and covers a displacement half of its amplitude in a time '$t$', the further time taken by it to reach mean position is

physics simple harmonic motion representing shm with circular motion shm as projection of circular motion simple harmonic motion (shm) as a projection of uniform circular motion

  1. $2t$

  2. $t$

  3. $ t/\sqrt{2} $

  4. $t/2$

Show answer
Correct Option: D
Explanation:

$x=A cos(\omega t)$
Taking $t=0$ at extreme position.
$x=\dfrac{A}{2}$ is reached in time t
$\omega t=\dfrac{\pi }{3} ;\omega =\dfrac{\pi }{3t}$
$x=0$ is reached at $\omega t _{1}=\dfrac{\pi }{2}$
$t _{1}=\dfrac{\pi }{2\omega }=\dfrac{\pi \times 3t}{2\times\pi }=\dfrac{3t}{2}$
$t _{1}-t=\dfrac{t}{2}$

The circular motion of a particle whose speed is constant is

physics simple harmonic motion representing shm with circular motion shm as projection of circular motion simple harmonic motion (shm) as a projection of uniform circular motion

  1. Periodic but not simple harmonic

  2. Simple harmonic but not periodic

  3. Periodic and simple harmonic

  4. Neither periodic not simple harmonic

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Correct Option: A
Explanation:

Uniform circular motion is a periodic motion but not simple harmonic

The correct option is (a)

Find the distance covered by a particle from time $t=0$ to $t=6\ \sec$. When the particle followsa the movement in straight line according to $y=a\cos \left(\dfrac {\pi}{4}\right)t$:-

physics simple harmonic motion representing shm with circular motion shm as projection of circular motion simple harmonic motion (shm) as a projection of uniform circular motion

  1. $a$

  2. $2a$

  3. $3a$

  4. $4a$

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Correct Option: C

The particle executes SHM on a straight line. At two positions its velocity $u$ and $v$ while acceleration, $\alpha$ and $\beta$ respectively $[\beta > \alpha >0]$, the distance between the two positions will be:-

physics simple harmonic motion representing shm with circular motion shm as projection of circular motion simple harmonic motion (shm) as a projection of uniform circular motion

  1. $\dfrac{u^2+v^2}{\alpha+\beta}$

  2. $-\dfrac{u^2-v^2}{\alpha+\beta}$

  3. $\dfrac{u^2-v^2}{\alpha-\beta}$

  4. $\dfrac{u^2+v^2}{\beta-\alpha}$

Show answer
Correct Option: B
Explanation:

We know that velocity is related as,

$u=\omega \sqrt{A^2-X _1^2}$
$v=\omega\sqrt{A^2-X _2^2}$
Acceleration is related as,
$\alpha=-\omega^2X _1^2$
$\beta=-\omega^2X _2^2$
$\dfrac{u^2-v^2}{u^2}=-(X _1^2-X _2^2)$
$=-(X _1+X _2)(X _1-X _2)$
$=\dfrac{\alpha+\beta}{u^2}(X _1-X _2)$
$X _1-X _2=-\dfrac{u^2-v^2}{\alpha+\beta}$

A ball is projected from the ground at angle 0 with the horizontal. After 1 sec it is moving at angle ${ 45 }^{ \circ  }$ with the horizontal and after 2s it is moving horizontally. What is the velocity of projection of the ball ? (Take $g=10\quad { ms }^{ -2 }$)

physics simple harmonic motion representing shm with circular motion shm as projection of circular motion simple harmonic motion (shm) as a projection of uniform circular motion

  1. $10\sqrt { { 3ms }^{ -1 } } $

  2. $20\sqrt { { 3ms }^{ -1 } } $

  3. $10\sqrt { { 5ms }^{ -1 } } $

  4. $20\sqrt { { 2ms }^{ -1 } } $

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Correct Option: C

To understand Simple Harmonic Motion as analogous to circular motion,

physics simple harmonic motion representing shm with circular motion shm as projection of circular motion simple harmonic motion (shm) as a projection of uniform circular motion

  1. we project the circular motion of the particle along any radius.

  2. we project the circulation motion of the particle along a chord.

  3. we project the circulation motion of the particle along the diameter.

  4. None of these.

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Correct Option: C
Explanation:

The circular motion can be resolved into a S.H.M. In this process we project circular motion of particle along diameter the projection of particle performs S.H.M. with centres as mean positions.

The actual distance moved along the circle will be                   the distance moved by the projection on the diameter.

physics simple harmonic motion representing shm with circular motion shm as projection of circular motion simple harmonic motion (shm) as a projection of uniform circular motion

  1. less than

  2. equal to

  3. greater than

  4. None of these

Show answer
Correct Option: C
Explanation:

Let the radius of the circle be $R$.

Thus distance moved along the circle  $D = \pi R = 3.14 R$
Projection of this distance along the diameter is equal to the diameter of the circle i.e.  $2R$
Thus actual distance moved along the circle will be greater than the distance moved by the projection on the diameter.