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Simple Harmonic Motion

JEE Physics · 108 questions · Page 2 of 11 · Click an option or "Show Solution" to reveal answer

Q11
The total energy of particle, executing simple harmonic motion is
A independent of xx
B ∝ x2 \propto \,{x^2}
C ∝ x \propto \,x
D ∝ x1/2 \propto \,{x^{1/2}}
Correct Answer
Option A
Solution

At any instant the total energy is

12kA2=  {1 \over 2}k{A^2} = \,\,

constant, where

A=A=

amplitude hence total energy is independent of

x.x.
Q12
The motion of a simple pendulum executing S.H.M. is represented by the following equation. y=Asin⁡(πt+ϕ)y = A\sin (\pi t + \phi ), where time is measured in second. The length of pendulum is
A 97.23 cm
B 25.3 cm
C 99.4 cm
D 406.1 cm
Correct Answer
Option C
Solution
ω=π=gl\omega = \pi = \sqrt {{g \over l}}

So,

l=gπ2l = {g \over {{\pi ^2}}}
≃99.4\simeq 99.4

cm (Nearest value)

Q13
The bob of a simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation, till water is coming out, the time period of oscillation would
A first decrease and then increase to the original value
B first increase and then decrease to the original value
C increase towards a saturation value
D remain unchanged
Correct Answer
Option B
Solution

Center of mass of combination of liquid and hollow portion (at position

ℓ\ell

), first goes down (to

ℓ+Δℓ\ell + \Delta \ell

) and when total water is drained out, center of mass regain its original position (to

ℓ\ell

),

T=2πℓgT = 2\pi \sqrt {{\ell \over g}}

\therefore$

′T′'T'

first increases and then decreases to original value.

Q14
The function sin⁡2(ωt){\sin ^2}\left( {\omega t} \right) represents
A a periodic, but not SHMSHM with a period πω{\pi \over \omega }
B a periodic, but not SHMSHM with a period 2πω{{2\pi } \over \omega }
C a SHMSHM with a period πω{\pi \over \omega }
D a SHMSHM with a period 2πω{{2\pi } \over \omega }
Correct Answer
Option A
Solution

y = sin2ω\omegat =

1−cos⁡2ωt2{{1 - \cos 2\omega t} \over 2}
=12−12cos⁡ 2ωt= {1 \over 2} - {1 \over 2}\cos \,2\omega t

∴\therefore Angular speed = 2ω\omega ∴\therefore Period (T) =

2πangular speed{{2\pi } \over {angular\,speed}}

=

2π2ω{{2\pi } \over {2\omega }}

=

πω{\pi \over \omega }

So it is a periodic function. As y = sin2ω\omegat

dydt{{dy} \over {dt}}

= 2ω\omegasinω\omegat cosω\omegat = ω\omega sin2ω\omegat

d2ydt2{{{d^2}y} \over {d{t^2}}}

=

2ω22{\omega ^2}

cos2ω\omegat which is not proportional to -y. Hence it is is not SHM.

Q15
A coin is placed on a horizontal platform which undergoes vertical simple harmonic motoin of angular frequency ω.\omega . The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time
A at the mean position of the platform
B for an amplitude of gω2{g \over {{\omega ^2}}}
C For an amplitude of g2ω2{{{g^2}} \over {{\omega ^2}}}
D at the height position of the platform
Correct Answer
Option B
Solution

Coin

AA

is moving in simple harmonic motion in vertical direction.

Now we are assuming coin will leave contact with the platform when platform is at a distance of

xx

from the mean position which is also called amplitude. At distance

xx

the force acting on the coin is

mg−N=mω2xmg - N = m{\omega ^2}x

For coin to leave contact

N=0N=0
⇒mg=mω2x⇒x=gω2\Rightarrow mg = m{\omega ^2}x \Rightarrow x = {g \over {{\omega ^2}}}

∴\therefore Option (B) is correct.

Q16
Starting from the origin a body oscillates simple harmonically with a period of 22 s.s. After what time will its kinetic energy be 75%75\% of the total energy?
A 16s{1 \over 6}s
B 14s{1 \over 4}s
C 13s{1 \over 3}s
D 112s{1 \over 12}s
Correct Answer
Option A
Solution
K.E. K.E.\,

of a body undergoing

SHMSHM

is given by,

K.E.=12ma2ω2cos⁡2 ωt,K.E. = {1 \over 2}m{a^2}{\omega ^2}{\cos ^2}\,\omega t,
T.E.=12ma2ω2T.E. = {1 \over 2}m{a^2}{\omega ^2}

Given

K.E.=0.75T.E.K.E.=0.75T.E.
⇒0.75=cos⁡2ωt⇒ωt=π6\Rightarrow 0.75 = {\cos ^2}\omega t \Rightarrow \omega t = {\pi \over 6}
⇒t=π6×ω⇒t=π×26×2π⇒t=16s\Rightarrow t = {\pi \over {6 \times \omega }} \Rightarrow t = {{\pi \times 2} \over {6 \times 2\pi }} \Rightarrow t = {1 \over 6}s
Q17
A point mass oscillates along the xx-axis according to the law x=x0 cos⁡(ωt−π/4).x = {x_0}\,\cos \left( {\omega t - \pi /4} \right). If the acceleration of the particle is written as a=A cos⁡(ωt+δ),a = A\,\cos \left( {\omega t + \delta } \right), then
A A=x0ω2,  δ=3π/4A = {x_0}{\omega ^2},\,\,\delta = 3\pi /4
B A=x0,  δ=−π/4A = {x_0},\,\,\delta = - \pi /4
C A=x0ω2,  δ=π/4A = {x_0}{\omega ^2},\,\,\delta = \pi /4
D A=x0ω2,  δ=−π/4A = {x_0}{\omega ^2},\,\,\delta = - \pi /4
Correct Answer
Option A
Solution

Here,

x=x0cos⁡(ωt−π/4)x = {x_0}\cos \left( {\omega t - \pi /4} \right)

∴\therefore Velocity,

v=dxdt=−x0ωsin⁡(ωt−π4)v = {{dx} \over {dt}} = - {x_0}\omega \sin \left( {\omega t - {\pi \over 4}} \right)

Acceleration,

a=dvdt=−x0ω2cos⁡(ωt−π4)a = {{dv} \over {dt}} = - {x_0}{\omega ^2}\cos \left( {\omega t - {\pi \over 4}} \right)
=x0ω2cos⁡[π+(ωt−π4)]= {x_0}{\omega ^2}\cos \left[ {\pi + \left( {\omega t - {\pi \over 4}} \right)} \right]
=x0ω2cos⁡(ωt+3π4)= {x_0}{\omega ^2}\cos \left( {\omega t + {{3\pi } \over 4}} \right)
           ...(1)\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)

Acceleration,

a=Acos⁡(ωt+δ)a = A\cos \left( {\omega t + \delta } \right)
           ...(2)\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)

Comparing the two equations, we get

A=x0ω2A = {x_0}{\omega ^2}

and

δ=3π4.\delta = {{3\pi } \over 4}.
Q18
If x,x, vv and aa denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period T,T, then, which of the following does not change with time?
A aT/xaT/x
B aT+2πvaT + 2\pi v
C aT/vaT/v
D a2T2+4π2v2{a^2}{T^2} + 4{\pi ^2}{v^2}
Correct Answer
Option A
Solution

For an

SHM,SHM,

the acceleration

a=−ω2xa = - {\omega ^2}x

where

ω2{\omega ^2}

is a constant. Therefore

ax{a \over x}

is a constant. The time period

TT

is also constant. Therefore

aTx{{aT} \over x}

is a constant.

Q19
Two particles are executing simple harmonic motion of the same amplitude AA and frequency ω\omega along the xx-axis. Their mean position is separated by distance X0(X0>A){X_0}\left( {{X_0} > A} \right). If the maximum separation between them is (X0+A),\left( {{X_0} + A} \right), the phase difference between their motion is:
A π3{\pi \over 3}
B π4{\pi \over 4}
C π6{\pi \over 6}
D π2{\pi \over 2}
Correct Answer
Option A
Solution

We know that, equation for SHM along x-axis is given by

x=Asin⁡(ωt+ϕ)x = A\sin (\omega t + \phi )

Let mean position for 1st particle is at x = 0 So, the SHM equation for 1st particle,

x1=Asin⁡(ωt+ϕ1){x_1} = A\sin (\omega t + {\phi _1})

Now, as the separation between mean positions of both the particle is x0_0 So, mean position for 2nd particle is x=x0x=x_0 Hence, the SHM equation for 2nd particle,

x2=x0+Asin⁡(ωt+ϕ2){x_2} = {x_0} + A\sin (\omega t + {\phi _2})

Now, the separation between particles would be

∣x2−x1∣=x0+Asin⁡(ωt+ϕ2)−Asin⁡(ωt+ϕ1)\left| {{x_2} - {x_1}} \right| = {x_0} + A\sin (\omega t + {\phi _2}) - A\sin (\omega t + {\phi _1})
=x0+A[sin⁡(ωt+ϕ2)−sin⁡(ωt+ϕ1)]= {x_0} + A\left[ {\sin (\omega t + {\phi _2}) - \sin (\omega t + {\phi _1})} \right]

We know,

sin⁡C−sin⁡D=2cos⁡(C+D2)sin⁡(C−D2)\sin C - \sin D = 2\cos \left( {{{C + D} \over 2}} \right)\sin \left( {{{C - D} \over 2}} \right)

Hence,

∣x2−x1∣=x0+2Acos⁡(ωt+ϕ1+ϕ22)sin⁡(ϕ2−ϕ12)\left| {{x_2} - {x_1}} \right| = {x_0} + 2A\cos \left( {\omega t + {{{\phi _1} + {\phi _2}} \over 2}} \right)\sin \left( {{{{\phi _2} - {\phi _1}} \over 2}} \right)

For maximum separation, Let

cos⁡(ωt+ϕ1+ϕ22)=1\cos \left( {\omega t + {{{\phi _1} + {\phi _2}} \over 2}} \right) = 1

So,

∣x2−x1∣=x0+2Asin⁡(ϕ2−ϕ12)\left| {{x_2} - {x_1}} \right| = {x_0} + 2A\sin \left( {{{{\phi _2} - {\phi _1}} \over 2}} \right)

Since,

∣x2−x1∣=x0+A\left| {{x_2} - {x_1}} \right| = {x_0} + A

(given) So,

x0+A=x0+2Asin⁡(ϕ2−ϕ12){x_0} + A = {x_0} + 2A\sin \left( {{{{\phi _2} - {\phi _1}} \over 2}} \right)
⇒sin⁡(ϕ2−ϕ12)=12\Rightarrow \sin \left( {{{{\phi _2} - {\phi _1}} \over 2}} \right) = {1 \over 2}
⇒ϕ2−ϕ12=π6\Rightarrow {{{\phi _2} - {\phi _1}} \over 2} = {\pi \over 6}
⇒ϕ2−ϕ1=π3\Rightarrow {\phi _2} - {\phi _1} = {\pi \over 3}
Q20
A mass M,M, attached to a horizontal spring, executes S.H.M.S.H.M. with amplitude A1.{A_1}. When the mass MM passes through its mean position then a smaller mass mm is placed over it and both of them move together with amplitude A2.{A_2}. The ratio of (A1A2)\left( {{{{A_1}} \over {{A_2}}}} \right) is :
A M+mM{{M + m} \over M}
B (MM+m)12{\left( {{M \over {M + m}}} \right)^{{1 \over 2}}}
C (M+mM)12{\left( {{{M + m} \over M}} \right)^{{1 \over 2}}}
D MM+m{M \over {M + m}}
Correct Answer
Option C
Solution

The net force becomes zero at the mean point. Therefore, linear momentum must be conserved. ∴\therefore

Mv1=(M+m)v2M{v_1} = \left( {M + m} \right){v_2}
MA1kM=(M+m)A2km+MM{A_1}\sqrt {{k \over M}} = \left( {M + m} \right){A_2}\sqrt {{k \over {m + M}}}

∴\therefore

(V=AkM)\left( {V = A\sqrt {{k \over M}} } \right)
A1M=A2M+m{A_1}\sqrt M = {A_2}\sqrt {M + m}

∴\therefore

A1A2=m+MM{{{A_1}} \over {{A_2}}} = \sqrt {{{m + M} \over M}}
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