Practice Start Test

Simple Harmonic Motion

NEET Physics · 94 questions · Page 3 of 10 · Click an option or "Show Solution" to reveal answer

Q21
A particle is executing a simple harmonic motion. Its maximum acceleration is α\alpha and maximum velocity is β\beta . Them, its time period of vibration will be
A β2α{{{\beta ^2}} \over \alpha }
B 2πβα{{2\pi \beta } \over \alpha }
C β2α2{{{\beta ^2}} \over {{\alpha ^2}}}
D αβ{\alpha \over \beta }
Correct Answer
Option B
Solution

As, we know, in SHM Maximum acceleration of the particle, α\alpha = A

ω2\omega ^2

Maximum velocity, β\beta = Aω\omega

⇒ω=αβ\Rightarrow \omega = {\alpha \over \beta }
⇒T=2πω=2πβα\Rightarrow T = {{2\pi } \over \omega } = {{2\pi \beta } \over \alpha }
[∵ω=2πT]\left[ \because {\omega = {{2\pi } \over T}} \right]
Q22
When two displacements represented by y 1 = a sin(ωt){\left( {\omega t} \right)} and y 2 = b cos(ωt){\left( {\omega t} \right)} aresuperimposed the motion is
A simple harmonic with amplitude a2+b2\sqrt {{a^2} + {b^2}}
B simple harmonic with amplitude (a+b)2{{\left( {a + b} \right)} \over 2}
C not a simple harmonic
D simple harmonic with amplitude ab{a \over b}
Correct Answer
Option A
Solution

Here,

y1=asin⁡ωt{y_1} = a\sin \omega t
y2=bcos⁡ω=bsin⁡(ωt+π2){y_2} = b\cos \omega = b\sin \left( {\omega t + {\pi \over 2}} \right)

Hence, resultant motion is SHM with amplitude

a2+b2\sqrt {{a^2} + {b^2}}

.

Q23
A particle is executing SHM along a straight line. Its velocities at distances x 1 and x 2 from the mean position are V 1 and V 2 respectively. Its time period is
A 2πV12+V22x12+x222\pi \sqrt {{{V_1^2 + V_2^2} \over {x_1^2 + x_2^2}}}
B 2πV12−V22x12−x222\pi \sqrt {{{V_1^2 - V_2^2} \over {x_1^2 - x_2^2}}}
C 2πx12+x22V12+V222\pi \sqrt {{{x_1^2 + x_2^2} \over {V_1^2 + V_2^2}}}
D 2πx22−x12V12−V222\pi \sqrt {{{x_2^2 - x_1^2} \over {V_1^2 - V_2^2}}}
Correct Answer
Option D
Solution

As we know, for particle undergoing SHM,

V=ωA2−X2V = \omega \sqrt {{A^2} - {X^2}}
V12=ω2(A2−x12)V_1^2 = {\omega ^2}\left( {{A^2} - x_1^2} \right)
V22=ω2(A2−x22)V_2^2 = {\omega ^2}\left( {{A^2} - x_2^2} \right)

Substracting we get,

V12ω2+x12=V22ω2+x22{{V_1^2} \over {{\omega ^2}}} + x_1^2 = {{V_2^2} \over {{\omega ^2}}} + x_2^2
⇒V12−V22ω2=x22−x12\Rightarrow {{V_1^2 - V_2^2} \over {{\omega ^2}}} = x_2^2 - x_1^2
⇒w=V12−V22x22−x12\Rightarrow w = \sqrt {{{V_1^2 - V_2^2} \over {x_2^2 - x_1^2}}}
⇒T=2πx22−x12V12−V22\Rightarrow T = 2\pi \sqrt {{{x_2^2 - x_1^2} \over {V_1^2 - V_2^2}}}
Q24
A particle of mass m oscillates along x-axis according to equation x = asinω\omega t. The nature of the graph between momentum and displacement of the particle is
A Circle
B Hyperbola
C Ellipse
D Straight line passing through origin
Correct Answer
Option C
Solution

As

v2a2ω2+y2a2=1{{{v^2}} \over {{a^2}{\omega ^2}}} + {{{y^2}} \over {{a^2}}} = 1

, This is the equation of ellipse. Hence the graph is an ellipse. P versus x graph is similar to V versus x graph.

Q25
Two particles are oscillating along two close parallel straight lines side by side, with the same frequency and amplitudes. They pass each other, moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to the paths of the two particles. The phase difference is
A π6{\pi \over 6}
B 0
C 2π3{{2\pi } \over 3}
D π\pi
Correct Answer
Option C
Solution

Equation of SHM is given by

x=Asin⁡(ωt+δ)x = A\sin \left( {\omega t + \delta } \right)
(ωt+δ)\left( {\omega t + \delta } \right)

is called phase. When x =

A2{A \over 2}

, then

sin⁡(ωt+δ)=12\sin \left( {\omega t + \delta } \right) = {1 \over 2}
⇒ωt+δ=π6{ \Rightarrow \omega t + \delta = {\pi \over 6}}
⇒ϕ1=π6\Rightarrow {\phi _1} = {\pi \over 6}

For second particle,

ϕ2=π−π6=5π6{\phi _2} = \pi - {\pi \over 6} = {{5\pi } \over 6}

∴\therefore

ϕ=ϕ2−ϕ1\phi = {\phi _2} - {\phi _1}
=4π6=2π3= {{4\pi } \over 6} = {{2\pi } \over 3}
Q26
Out of the following functions representing motion of a particle which represents SHM (1) y = sinω\omega t −- cosω\omega t (2) y = sin 3 ω\omega t (3) y = 5cos(3π4−3ωt)\left( {{{3\pi } \over 4} - 3\omega t} \right) (4) y = 1 + ω\omega t + ω\omega 2 t 2
A Only (1)
B Only (4) does not represent SHM
C Only (1) and (3)
D Only (1) and (2)
Correct Answer
Option C
Solution

y = sinω\omegat – cosω\omegat

=2[12sin⁡ωt−12cos⁡ωt]= \sqrt 2 \left[ {{1 \over {\sqrt 2 }}\sin \omega t - {1 \over {\sqrt 2 }}\cos \omega t} \right]
=2sin⁡(ωt−π4)= \sqrt 2 \sin \left( {\omega t - {\pi \over 4}} \right)

It represents a SHM with time period,

T=2πωT = {{2\pi } \over \omega }
y=sin⁡3ωt=14[3sin⁡ωt−sin⁡3ωt]y = {\sin ^3}\omega t = {1 \over 4}\left[ {3\sin \omega t - \sin 3\omega t} \right]

It represents a periodic motion with time period

T=2πωT = {{2\pi } \over \omega }

but now SHM.

y=5cos⁡(3π4−3ωt)y = 5\cos \left( {{{3\pi } \over 4} - 3\omega t} \right)
=5cos⁡(3ωt−3π4)= 5\cos \left( {3\omega t - {{3\pi } \over 4}} \right)
[∵cos⁡(−θ)=cos⁡θ]\left[ \because {\cos \left( { - \theta } \right) = \cos \theta } \right]

It represents a SHM with time period,

T=2π3ωT = {{2\pi } \over {3\omega }}
y=1+ωt+ω2t2y = 1 + \omega t + {\omega ^2}{t^2}

It represents a non-periodic motion. Also it is not physically acceptable as y →\to ∞\infty as t →\to ∞\infty.

Q27
The displacement of a particle along the x-axis is given by x = asin 2 ω\omega t. The motion of the particle corresponds to
A simple harmonic motion of frequency ω\omega /π\pi
B simple harmonic motion of frequency 3ω/2π3\omega /2\pi
C non simple harmonic motion
D simple harmonic motion of frequency ω/2π\omega /2\pi
Correct Answer
Option C
Solution
x=asin⁡2ωt=a(1−cos⁡2ωt2)x = a{\sin ^2}\omega t = a\left( {{{1 - \cos 2\omega t} \over 2}} \right)
(∵cos⁡2θ=1−2sin⁡2θ)\left(\because {\cos 2\theta = 1 - 2{{\sin }^2}\theta } \right)
=a2−acos⁡2ωt2= {a \over 2} - {{a\cos 2\omega t} \over 2}

∴\therefore Velocity,

v=dxdt=2ωasin⁡2ωt2=ωαsin⁡2ωtv = {{dx} \over {dt}} = {{2\omega a\sin 2\omega t} \over 2} = \omega \alpha \sin 2\omega t

Acceleration,

a=dvdt=2ω2acos⁡2ωta = {{dv} \over {dt}} = 2{\omega ^2}a\cos 2\omega t

For the given displacement x =

αsin⁡2ωt\alpha \sin 2\omega t

,

a∝−xa \propto - x

is not satisfied.

Hence, the motion of the particle is non simple harmonic motion.

Note : The given motion is a periodic motion with a time period

T=2π2ω=πωT = {{2\pi } \over {2\omega }} = {\pi \over \omega }
Q28
The period of oscillation of a mass M suspended from a strong of negligible mass is T. If along with it another mass M is also suspended, the period of oscillation will now be
A T
B T2{T \over {\sqrt 2 }}
C 2T
D 2T\sqrt 2 T
Correct Answer
Option D
Solution
T=2πmKT = 2\pi \sqrt {{m \over K}}
∴T1T2=M1M2\therefore {{{T_1}} \over {{T_2}}} = \sqrt {{{{M_1}} \over {{M_2}}}}

∴\therefore

T2=T1M2M1=T12MM{T_2} = {T_1}\sqrt {{{{M_2}} \over {{M_1}}}} = {T_1}\sqrt {{{2M} \over M}}
T2=T12=2T{T_2} = {T_1}\sqrt 2 = \sqrt 2 T

(where T 1 =T)

Q29
Which one of the following equations of motion represents simple harmonic motion ? where k, k 0 , k 1 and a are all positive.
A Acceleration = −- k (x + a)
B Acceleration = k(x + a)
C Acceleration = −- ω\omega 2 x
D Acceleration = −- k 0 x + k 1 x 2
Correct Answer
Option C
Solution

Simple harmonic motion is defined as follows Acceleration

d2ydt2=−ω2x{{{d^2}y} \over {d{t^2}}} = - {\omega ^2}x

The negative sign is very important in simple harmonic motion.

Acceleration is independent of any initial displacement of equilibrium position.

Then acceleration =

−ω2x- {\omega ^2}x

.

Q30
A simple pendulum performs simple harmonic motion about x = 0 with an amplitude a and time period T. The speed of the pendulum at x = a/2 will be
A πaT{{\pi a} \over T}
B 3π2aT{{3{\pi ^2}a} \over T}
C πa3T{{\pi a\sqrt 3 } \over T}
D πa32T{{\pi a\sqrt 3 } \over {2T}}
Correct Answer
Option C
Solution

Speed

v=ωa2−x2,x=a2v = \omega \sqrt {{a^2} - {x^2}} ,x = {a \over 2}

∴\therefore

v=ωa2−a24=ω3a24v = \omega \sqrt {{a^2} - {{{a^2}} \over 4}} = \omega \sqrt {{{3{a^2}} \over 4}}
=2πTa32=πa3T= {{2\pi } \over T}{{a\sqrt 3 } \over 2} = {{\pi a\sqrt 3 } \over T}
Ready for a full NEET mock test? Timed · full syllabus · instant results
Take a Mock Test →