12: Simple Harmonic Motion : Cbse-xi-physics

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CBSE-XI-Physics

12: Simple Harmonic Motion

with Solutions - page 5

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Qstn #3
A particle executes simple harmonic motion with an amplitude of 10 cm. At what distance from the mean position are the kinetic and potential energies equal?
Ans : It is given that:
Amplitude of the particle executing simple harmonic motion, A = 10 cm
To determine the distance from the mean position, where the kinetic energy of the particle is equal to its potential energy:
Let y be displacement of the particle,
`` \omega `` be the angular speed of the particle, and
A be the amplitude of the simple harmonic motion.
Equating the mathematical expressions for K.E. and P.E. of the particle, we get:
`` \left(\frac{1}{2}\right)m{\,\mathrm{\,\omega \,}}^{2}\left({A}^{2}-{y}^{2}\right)=\left(\frac{1}{2}\right)m{\,\mathrm{\,\omega \,}}^{2}{y}^{2}``
A² - y² = y²
2y² = A²
`` \Rightarrow y=\frac{A}{\sqrt{2}}=\frac{10}{\sqrt{2}}=5\sqrt{2}``
The kinetic energy and potential energy of the particle are equal at a distance of `` 5\sqrt{2}`` cm from the mean position.
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Qstn #4
The maximum speed and acceleration of a particle executing simple harmonic motion are 10 cm s^-1 and 50 cm s^-2. Find the position(s) of the particle when the speed is 8 cm s^-1.
Ans : It is given that:
Maximum speed of the particle, `` {v}_{Max}`` = 10 cms^{`` -1``}
Maximum acceleration of the particle, `` {a}_{Max}`` = 50 cms^-2
The maximum velocity of a particle executing simple harmonic motion is given by,
`` {v}_{Max}=A\omega ``
where `` \omega \,\mathrm{\,is\,}\,\mathrm{\,angular\,}\,\mathrm{\,frequency\,},\,\mathrm{\,and\,}``
A is amplitude of the particle.
Substituting the value of `` {v}_{Max}`` in the above expression, we get:
Aω = 10 `` ...\left(1\right)``
`` \Rightarrow {\,\mathrm{\,\omega \,}}^{2}=\frac{100}{{A}^{2}}``
a_Max = ω²A = 50 cms^-1
`` \Rightarrow {\omega }^{2}=\frac{50}{A}...\left(2\right)``
`` \,\mathrm{\,From\,}\,\mathrm{\,the\,}\,\mathrm{\,equations\,}\left(1\right)\,\mathrm{\,and\,}\left(2\right),\,\mathrm{\,we\,}\,\mathrm{\,get\,}:``
`` \frac{100}{{A}^{2}}=\frac{50}{A}``
`` \Rightarrow A=2\,\mathrm{\,cm\,}``
`` \mathbf{\therefore }\omega =\sqrt{\frac{100}{{A}^{2}}}=5{\,\mathrm{\,sec\,}}^{-1}``
To determine the positions where the speed of the particle is 8 ms^-1, we may use the following formula:
v² = ω² (A² - y²)
where y is distance of particle from the mean position, and
v is velocity of the particle.
On substituting the given values in the above equation, we get:
64 = 25 (4 - y²)
`` \Rightarrow \frac{64}{25}=4-{y}^{2}``
⇒ 4 - y² = 2.56
⇒ y² = 1.44
⇒ y = `` \sqrt{1.44}``
⇒ y = ± 1.2 cm (from the mean position)
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Qstn #5
A particle having mass 10 g oscillates according to the equation x = (2.0 cm) sin [(100 s^-1)t + π/6]. Find
Ans : Given:
Equation of motion of the particle executing S.H.M.,
`` x=\left(2.0\,\mathrm{\,cm\,}\right)\,\mathrm{\,sin\,}\left[\left(100{s}^{-1}\right)t+\frac{\,\mathrm{\,\pi \,}}{6}\right]``
`` \,\mathrm{\,Mass\,}\,\mathrm{\,of\,}\,\mathrm{\,the\,}\,\mathrm{\,particle\,},m=10\,\mathrm{\,g\,}`` ...(1)
General equation of the particle is given by,
`` x=A\,\mathrm{\,sin\,}(\omega t+\varphi )`` ...(2)
On comparing the equations (1) and (2) we get:

#5-a
the amplitude, the time period and the spring constant.
Ans : Amplitude, A is 2 cm.
Angular frequency, ω is 100 s^-1.
`` \,\mathrm{\,Time\,}\,\mathrm{\,period\,}\,\mathrm{\,is\,}\,\mathrm{\,calculated\,}\,\mathrm{\,as\,},``
`` T=\frac{2\,\mathrm{\,\pi \,}}{\omega }=\frac{2\,\mathrm{\,\pi \,}}{100}=\frac{\,\mathrm{\,\pi \,}}{50}\,\mathrm{\,s\,}``
`` =0.063\,\mathrm{\,s\,}``
Also, we know -
`` T=2\,\mathrm{\,\pi \,}\sqrt{\frac{m}{k}}``
`` \,\mathrm{\,where\,}\,\mathrm{\,k\,}\,\mathrm{\,is\,}\,\mathrm{\,the\,}\,\mathrm{\,sprin\,}g\,\mathrm{\,constant\,}.``
`` \Rightarrow {T}^{2}=4{\,\mathrm{\,\pi \,}}^{2}\frac{m}{k}``
`` \Rightarrow k=\frac{4{\,\mathrm{\,\pi \,}}^{2}m}{{T}^{2}}={10}^{5}\,\mathrm{\,dyne\,}/\,\mathrm{\,cm\,}``
`` =100\,\mathrm{\,N\,}/\,\mathrm{\,m\,}``

#5-b
the position, the velocity and the acceleration at t = 0.
Ans : At t = 0 and x = 2 cm `` \,\mathrm{\,sin\,}\frac{\,\mathrm{\,\pi \,}}{6}``
`` =2\times \frac{1}{2}=1\,\mathrm{\,cm\,}\,\mathrm{\,from\,}\,\mathrm{\,the\,}\,\mathrm{\,mean\,}\,\mathrm{\,position\,},``
We know:
x = A sin (ωt + ϕ)
Using `` v=\frac{dx}{dt},`` we get:
v = Aω cos (ωt + ϕ)
`` =2\times 100\,\mathrm{\,cos\,}\left(0+\frac{\,\mathrm{\,\pi \,}}{6}\right)``
`` =200\times \frac{\sqrt{3}}{2}``
`` =100\sqrt{3}{\,\mathrm{\,cms\,}}^{-1}``
`` =1.73{\,\mathrm{\,ms\,}}^{-1}``
Acceleration of the particle is given by,
a = `` -{\omega }^{2}``x
= 100²×1 = 10000 cm/s²
Page No 252:

Qstn #6
The equation of motion of a particle started at t = 0 is given by x = 5 sin (20t + π/3), where x is in centimetre and t in second. When does the particle
Ans : Given:
The equation of motion of a particle executing S.H.M. is,
`` x=5\,\mathrm{\,sin\,}\left(20t+\frac{\,\mathrm{\,\pi \,}}{3}\right)``
The general equation of S..H.M. is give by,
`` x=A\,\mathrm{\,sin\,}(\omega t+\varphi )``

#6-a
first come to rest
Ans : Maximum displacement from the mean position is equal to the amplitude of the particle.
As the velocity of the particle is zero at extreme position, it is at rest.
`` \therefore \,\mathrm{\,Displacement\,}`` x = 5, which is also the amplitude of the particle.
`` \Rightarrow 5=5\,\mathrm{\,sin\,}\left(20t+\frac{\,\mathrm{\,\pi \,}}{3}\right)``
`` \,\mathrm{\,Now\,},``
`` \,\mathrm{\,sin\,}\left(20t+\frac{\,\mathrm{\,\pi \,}}{3}\right)=1=\,\mathrm{\,sin\,}\frac{\,\mathrm{\,\pi \,}}{2}``
`` \Rightarrow 20t+\frac{\,\mathrm{\,\pi \,}}{3}=\frac{\,\mathrm{\,\pi \,}}{2}``
`` \Rightarrow t=\frac{\,\mathrm{\,\pi \,}}{120}\,\mathrm{\,s\,}``
The particle will come to rest at `` \frac{\,\mathrm{\,\pi \,}}{120}\,\mathrm{\,s\,}``

#6-b
first have zero acceleration
Ans : Acceleration is given as,
a = ω²x
`` ={\,\mathrm{\,\omega \,}}^{2}\left[5\,\mathrm{\,sin\,}\left(20t+\frac{\,\mathrm{\,\pi \,}}{3}\right)\right]``
For a = 0,
`` 5\,\mathrm{\,sin\,}\left(20t+\frac{\,\mathrm{\,\pi \,}}{3}\right)=0``
`` \Rightarrow \,\mathrm{\,sin\,}\left(20t+\frac{\,\mathrm{\,\pi \,}}{3}\right)=\,\mathrm{\,sin\,}\pi ``
`` \Rightarrow 20t=\,\mathrm{\,\pi \,}-\frac{\,\mathrm{\,\pi \,}}{3}=\frac{2\,\mathrm{\,\pi \,}}{3}``
`` \Rightarrow t=\frac{\,\mathrm{\,\pi \,}}{30}\,\mathrm{\,s\,}``

#6-c
first have maximum speed?
Ans : The maximum speed `` \left(v\right)`` is given by,
`` v=\,\mathrm{\,A\,}\omega \,\mathrm{\,cos\,}\left(\omega t+\frac{\,\mathrm{\,\pi \,}}{3}\right)`` (using `` v=\frac{dx}{dt}``)
`` =20\times 5\,\mathrm{\,cos\,}\left(20t+\frac{\,\mathrm{\,\pi \,}}{3}\right)``
For maximum velocity:
`` \,\mathrm{\,cos\,}\left(20t+\frac{\,\mathrm{\,\pi \,}}{3}\right)=-1=\,\mathrm{\,cos\,}\,\mathrm{\,\pi \,}``
`` \Rightarrow 20t=\,\mathrm{\,\pi \,}-\frac{\,\mathrm{\,\pi \,}}{3}=\frac{2\,\mathrm{\,\pi \,}}{3}``
`` \Rightarrow t=\frac{\,\mathrm{\,\pi \,}}{30}\,\mathrm{\,s\,}``
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Qstn #7
Consider a particle moving in simple harmonic motion according to the equation x = 2.0 cos (50 πt + tan^-1 0.75) where x is in centimetre and t in second. The motion is started at t = 0.
Ans : It is given that a particle executes S.H.M.
Equation of S.H.M. of the particle:
x = 2.0 cos (50`` \,\mathrm{\,\pi \,}``t + tan^-10.75)
= 2.0 cos (50`` \,\mathrm{\,\pi \,}``t + 0.643)

#7-a
When does the particle come to rest for the first time?
Ans : Velocity of the particle is given by,
`` v=\frac{dx}{dt}``
v = -100`` \,\mathrm{\,\pi \,}`` sin (50`` \,\mathrm{\,\pi \,}``t + 0.643)
As the particle comes to rest, its velocity becomes be zero.
⇒ v = -100`` \,\mathrm{\,\pi \,}`` sin (50`` \,\mathrm{\,\pi \,}``t + 0.643) = 0
⇒ sin (50`` \,\mathrm{\,\pi \,}``t + 0.643) = 0 = sin `` \,\mathrm{\,\pi \,}``
When the particle initially comes to rest,
50`` \,\mathrm{\,\pi \,}``t + 0.643 = `` \pi ``
⇒ t = 1.6 × 10^-2s

#7-b
When does he acceleration have its maximum magnitude for the first time?
Ans : Acceleration is given by,
`` a=\frac{dv}{dt}``
`` =-100\,\mathrm{\,\pi \,}\times 50\pi \,\mathrm{\,cos\,}\left(50\,\mathrm{\,\pi \,}t+0.643\right)``
For maximum acceleration:
cos (50`` \,\mathrm{\,\pi \,}``t + 0.643) = -1 = cos `` \,\mathrm{\,\pi \,}`` (max) (so that a is max)
⇒ t = 1.6 × 10^-2 s

#7-c
When does the particle come to rest for
Ans : When the particle comes to rest for the second time, the time is given as,
50`` \,\mathrm{\,\pi \,}``t + 0.643 = 2`` \,\mathrm{\,\pi \,}``
⇒ t = 3.6 × 10^-2 s
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Qstn #8
Consider a simple harmonic motion of time period T. Calculate the time taken for the displacement to change value from half the amplitude to the amplitude.
Ans : As per the conditions given in the question,
`` {y}_{1}=\frac{A}{2};``
`` {y}_{2}=A`` (for the given two positions)
Let y₁and y₂ be the displacements at the two positions and A be the amplitude.
Equation of motion for the displacement at the first position is given by,
y₁ = Asinωt₁
As displacement is equal to the half of the amplitude,
`` \frac{A}{2}=A\,\mathrm{\,sin\,}\omega {t}_{1}``
`` \Rightarrow \,\mathrm{\,sin\,}\omega {t}_{1}=\frac{1}{2}``
`` \Rightarrow \frac{2\,\mathrm{\,\pi \,}\times {t}_{1}}{T}=\frac{\,\mathrm{\,\pi \,}}{6}``
`` \Rightarrow {t}_{1}=\frac{T}{12}``
The displacement at second position is given by,
y₂ = A sin ωt₂
As displacement is equal to the amplitude at this position,
⇒ A = A sin ωt₂
⇒ sinωt₂ = 1
`` \Rightarrow \omega {t}_{2}=\frac{\,\mathrm{\,\pi \,}}{2}``
`` \Rightarrow \left(\frac{2\,\mathrm{\,\pi \,}}{\,\mathrm{\,T\,}}\right){t}_{2}=\frac{\,\mathrm{\,\pi \,}}{2}\left(\because \,\mathrm{\,sin\,}\frac{\,\mathrm{\,\pi \,}}{2}=1\right)``
`` \Rightarrow {t}_{2}=\frac{T}{4}``
`` ``
`` \therefore {t}_{2}-{t}_{1}=\frac{T}{4}-\frac{T}{12}=\frac{T}{6}``
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Qstn #9
The pendulum of a clock is replaced by a spring-mass system with the spring having spring constant 0.1 N m^-1. What mass should be attached to the spring?
Ans : Given:
Spring constant, k =0.1 N/m
Time period of the pendulum of clock, T = 2 s
Mass attached to the string, m, is to be found.
The relation between time period and spring constant is given as,
`` \,\mathrm{\,T\,}=2\,\mathrm{\,\pi \,}\sqrt{\left(\frac{m}{k}\right)}``
On substituting the respective values, we get:
`` 2=2\,\mathrm{\,\pi \,}\sqrt{\frac{m}{k}}``
`` \Rightarrow {\,\mathrm{\,\pi \,}}^{2}\left(\frac{m}{0.1}\right)=1``
`` \therefore m=\frac{0.1}{{\,\mathrm{\,\pi \,}}^{2}}=\frac{0.1}{10}``
`` =0.01\,\mathrm{\,kg\,}\approx 10\,\mathrm{\,g\,}``
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