12: Simple Harmonic Motion : Cbse-xi-physics

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CBSE-XI-Physics

12: Simple Harmonic Motion

with Solutions - page 2

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Qstn #14
A block of known mass is suspended from a fixed support through a light spring. Can you find the time period of vertical oscillation only by measuring the extension of the spring when the block is in equilibrium?
Ans : Yes.
Time period of a spring mass system is given by,
`` T=2\pi \sqrt{\frac{m}{k}}`` ...(1) .
where m is mass of the block, and
k is the spring constant
Time period is also given by the relation,
`` T=2\pi \sqrt{\frac{{x}_{0}}{g}}`` ...(2)
where, x₀ is extension of the spring, and
g is acceleration due to gravity
From the equations (1) and (2), we have:
`` mg=k{x}_{0}``
`` \Rightarrow k=\frac{mg}{{x}_{0}}``
Substituting the value of k in the above equation, we get:
`` T=2\pi \sqrt{\frac{m}{{\displaystyle \frac{mg}{{x}_{0}}}}}=2\,\mathrm{\,\pi \,}\sqrt{\frac{{\,\mathrm{\,x\,}}_{0}}{\,\mathrm{\,g\,}}}``
Thus, we can find the time period if the value of extension x₀ is known.
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Qstn #15
A platoon of soldiers marches on a road in steps according to the sound of a marching band. The band is stopped and the soldiers are ordered to break the steps while crossing a bridge. Why?
Ans : When the frequency of soldiers' feet movement becomes equal to the natural frequency of the bridge, and resonance occurs between soldiers' feet movement and movement of the bridge, maximum transfer of energy occurs from soldiers' feet to the bridge, which increases the amplitude of vibration. A continued increase in the amplitude of vibration, however, may lead to collapsing of the bridge.
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Qstn #16
The force acting on a particle moving along X-axis is F = -k(x - v_o t) where k is a positive constant. An observer moving at a constant velocity v₀ along the X-axis looks at the particle. What kind of motion does he find for the particle?
Ans : As the observer moves with a constant velocity along the same axis, he sees the same force on the particle and finds the motion of the particle is not simple harmonic motion.
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#
Section : ii

Qstn #1
A student says that he had applied a force
F=-kxon a particle and the particle moved in simple harmonic motion. He refuses to tell whether k is a constant or not. Assume that he was worked only with positive x and no other force acted on the particle.
(a) As x increases k increases.
(b) As x increases k decreases.
(c) As x increases k remains constant.
(d) The motion cannot be simple harmonic.
digAnsr: a
Ans : (a) As x increases k increases.
A body is said to be in simple harmonic motion only when,
F = `` -`` kx ...(1)
where F is force,
k is force constant, and
x is displacement of the body from the mean position.
Given:
F = -k`` \sqrt{x}`` ...(2)

On comparing the equations (1) and (2), it can be said that in order to execute simple harmonic motion, k should be proportional to `` \sqrt{x}`` . Thus, as x increases k increases.
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Qstn #2
The time period of a particle in simple harmonic motion is equal to the time between consecutive appearances of the particle at a particular point in its motion. This point is
(a) the mean position
(b) an extreme position
(c) between the mean position and the positive extreme
(d) between the mean position and the negative extreme
digAnsr: b
Ans : (b) an extreme position
One oscillation is said to be completed when the particle returns to the extreme position i.e. from where it started.
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Qstn #3
The time period of a particle in simple harmonic motion is equal to the smallest time between the particle acquiring a particular velocity
v→. The value of v is
(a) v_max
(b) 0
(c) between 0 and v_max
(d) between 0 and -v_max
digAnsr: a
Ans : (a) v_max
Because the time period of a simple harmonic motion is defined as the time taken to complete one oscillation_.
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Qstn #4
The displacement of a particle in simple harmonic motion in one time period is
(a) A
(b) 2A
(c) 4A
(d) zero
digAnsr: d
Ans : (d) zero
Displacement is defined as the distance between the starting and the end point through a straight line. In one complete oscillation, the net displacement is zero as the particle returns to its initial position.
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Qstn #5
The distance moved by a particle in simple harmonic motion in one time period is
(a) A
(b) 2A
(c) 4A
(d) zero
digAnsr: c
Ans : (c) 4A
In an oscillation, the particle goes from one extreme position to other extreme position that lies on the other side of mean position and then returns back to the initial extreme position. Thus, total distance moved by particle is,
2A + 2A = 4A.
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Qstn #6
The average acceleration in one time period in a simple harmonic motion is
(a) Aω²
(b) Aω²/2
(c)
Aω2/2
(d) zero
digAnsr: d
Ans : (d) zero
The acceleration changes its direction (to opposite direction) after every half oscillation. Thus, net acceleration is given as,
Aω²+ ( -Aω²) = 0
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Qstn #7
The motion of a particle is given by x = A sin ωt + B cos ωt. The motion of the particle is
(a) not simple harmonic
(b) simple harmonic with amplitude A + B
(c) simple harmonic with amplitude (A + B)/2
(d) simple harmonic with amplitude
A2+B2.
digAnsr: d
Ans : (d) simple harmonic with amplitude A sin ωt + B cos ωt ...(1)
`` \,\mathrm{\,Acceleration\,},``
`` a=\frac{{d}^{2}x}{d{\,\mathrm{\,t\,}}^{2}}=\frac{{d}^{2}}{d{\,\mathrm{\,t\,}}^{2}}(A\,\mathrm{\,sin\,}\omega t+B\,\mathrm{\,cos\,}\omega t)``
`` =\frac{\,\mathrm{\,d\,}}{\,\mathrm{\,dt\,}}(A\omega \,\mathrm{\,cos\,}\omega t-B\omega \,\mathrm{\,sin\,}\omega t)``
`` =-A{\omega }^{2}\,\mathrm{\,sin\omega t\,}-B{\omega }^{2}\,\mathrm{\,cos\,}\,\mathrm{\,\omega t\,}``
`` =-{\omega }^{2}(A\,\mathrm{\,sin\,}\omega t+B\,\mathrm{\,cos\,}\omega t)``
`` =-{\omega }^{2}x``
`` ``
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Qstn #8
The displacement of a particle is given by
r→=Ai→ cosωt + j→sinωt.The motion of the particle is
(a) simple harmonic
(b) on a straight line
(c) on a circle
(d) with constant acceleration
digAnsr: c
Ans : (c) on a circle
We know, `` \frac{{d}^{2}}{d{t}^{2}}\stackrel{\to }{r}=-{\omega }^{2}\stackrel{\to }{r}``
`` ``
But there is a phase difference of 90^o between the x and y components because of which the particle executes a circular motion and hence, the projection of the particle on the diameter executes a simple harmonic motion.
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Qstn #9
A particle moves on the X-axis according to the equation x = A + B sin ωt. The motion is simple harmonic with amplitude
(a) A
(b) B
(c) A + B
(d)
A2+B2.
digAnsr: b
Ans : (b) B
At t = 0,
Displacement `` \left({x}_{0}\right)`` is given by,
x₀ = A + sin ω(0) = A
Displacement x will be maximum when sinωt is 1
or,
x_m = A + B
Amplitude will be:
x_m`` -`` x_o = A + B `` -`` A = B
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Qstn #10
Figure represents two simple harmonic motions.

Figure
The parameter which has different values in the two motions is
(a) amplitude
(b) frequency
(c) phase
(d) maximum velocity
digAnsr: c
Ans : (c) phase
Because the direction of motion of particles A and B is just opposite to each other.
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Qstn #11
The total mechanical energy of a spring-mass system in simple harmonic motion is
E=12mω2A2.Suppose the oscillating particle is replaced by another particle of double the mass while the amplitude A remains the same. The new mechanical energy will
(a) become 2E
(b) become E/2
(c) become
2E
(d) remain E
digAnsr: d
Ans : (d) remain E
Mechanical energy (E) of a spring-mass system in simple harmonic motion is given by,
`` {E}_{}=\frac{1}{2}m{\omega }^{2}{A}^{2}``
where m is mass of body, and
`` \omega `` is angular frequency.
Let m₁ be the mass of the other particle and ω₁ be its angular frequency.
New angular frequency ω₁ is given by,
`` {\omega }_{1}=\sqrt{\frac{k}{{m}_{1}}}=\sqrt{\frac{k}{2m}}({m}_{1}=2m)``
New energy E₁ is given as,
`` {E}_{1}=\frac{1}{2}{m}_{1}{\omega }_{1}^{2}{A}^{2}``
`` =\frac{1}{2}\left(2m\right)(\sqrt{\frac{k}{2m}}{)}^{2}{A}^{2}``
`` =\frac{1}{2}m{\omega }^{2}{A}^{2}=E``
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