Q: 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 = \dfrac{a}{2} will be:

As simple pendulum performs simple harmonic motion. \therefore velocity, v = \omega \sqrt {a^{2} - x^{2}} At, x = \dfrac {a}{2} v = \dfrac {2\pi}{T} \sqrt {a^{2} -\left (\dfrac {a}{2}\right )^{2}} = \dfrac {2\pi}{T} \dfrac {\sqrt {3a^{2}}}{2} = \dfrac {\pi a\sqrt {3}}{T}

Q: The kinetic energy and potential energy of a particle executing SHM will be equal when displacement ( amplitude =a ) is

K.E.=P.E.\\ \Rightarrow \frac { 1 }{ 2 } m{ \omega }^{ 2 }({ a }^{ 2 }-{ x }^{ 2 })=\frac { 1 }{ 2 } m{ \omega }^{ 2 }{ x }^{ 2 }\\ \Rightarrow { a }^{ 2 }-{ x }^{ 2 }={ x }^{ 2 }\\ \Rightarrow 2{ x }^{ 2 }={ a }^{ 2 }\\ \Rightarrow x=a/\sqrt { 2 }

Question 1

Derive the equation for the kinetic energy and potential energy of a simple harmonic oscillator and show that the total energy of a particle in simple harmonic motion constant at any point on its path.

Accepted Answer

For SHM,

acceleration , a = -ω²y

F = ma = -mω²y

now, work , W = F.dy cos180° { because displacement and
acceleration are in opposite direction so, cos180° taken }

W = ∫mω²y.dy = mω²y²/2

use standard form of SHM , y = Asin(ωt ± Ф)

W = mω²A²/2 sin²(ωt ± Ф)

We know, Potential energy is work done stored in system .

so, P.E = W = mω²A²/2 sin²(ωt ± Ф)

again, Kinetic energy , K.E = 1/2mv² , here v is velocity

we know, v = ωAcos(ωt ± Ф)

so, K.E = mω²A²/2cos²(ωt ± Ф)

Total energy = K.E + P.E

= mω²A²/2 cos²(ωt ± Ф) + mω²A²/2
sin²(ωt ± Ф)

= mω²A²/2 [ cos²(ωt ± Ф) +
sin²(ωt ± Ф) ] = mω²A²/2 [ ∵sin²α + cos²α = 1]

= mω²A²/2 = constant

Question 2

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:

Accepted Answer

Using the
expressions for velocity for SHM, v_1 = \omega \sqrt{A^2
-x_1^2} v_2 = \omega \sqrt{A^2 -x_2^2} Squaring
both, v_1 ^2 = \omega ^2 ( A^2 -x_1 ^2)  ....(3) v_2 ^2 =
\omega ^2 ( A^2 -x_2 ^2)  ....(4)Subtracting (4) from (3),  v_1
^2 - v_2 ^2 = \omega ^2 ( x_2 ^2 - x_1 ^2)   \Rightarrow 
4 \pi ^2 ( x_2 ^2 - x_1 ^2) = T^2 (v_1 ^2 - v_2 ^2 ) 
\Rightarrow T = 2\pi \sqrt{\dfrac{ x_2 ^2 - x_1 ^2}{v_1 ^2 - v_2
^2}}  where we have used  \omega = \dfrac{2\pi}{T}

Question 3

A spring having with a spring constant 1200 N m \displaystyle ^{-1}  is mounted on a horizontal table as shown in the Figure. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass.

Accepted Answer

Spring constant,  k=1200\,m^{-1} Mass,  m=3kg Displacement,  A=2.0\,cm=0.02m (i) Frequency of oscillation v, is given by the relation: \displaystyle v=\frac{1}{T}=\frac{1}{2\pi}\sqrt{\frac{k}{m}} where, T is time period 	herefore \displaystyle v=\frac{1}{2	imes 3.14}\sqrt{\frac{1200}{3}} = 3.18 Hz Hence, the frequency of oscillations is 3.18 cycles per second.(ii) Maximum acceleration (a) is given by the relation: a=\omega^2\,A where, \omega =  Angular frequency  =\sqrt{\frac{k}{m}} A  = maximum displacement \displaystyle 	herefore a=\frac{k}{m}A=\frac{1200	imes 0.02}{3}=8\, ms^{-2} Hence, the maximum acceleration of the mass is  8.0\,m/s^2 (iii) Maximum velocity,  v_{max}=A\omega \displaystyle =A\sqrt{\frac{k}{m}}=0.02	imes \sqrt{\frac{1200}{3}}=0.4\,m/s Hence, the maximum velocity of the mass is 0.4 m/s.

Question 4

A particle executes linear simple harmonic motion with an amplitude of  3\ cm . When the particle is at  2\ cm  from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

Accepted Answer

Velocity of particle executing S.H.M  v=\omega \sqrt { { a }^{ 2 }-{ x }^{ 2 } }   and acceleration,  a=-{ \omega  }^{ 2 }x   x = displacement at any time interval a  = amplitude =3cm \omega = angular frequency,  =\cfrac { 2\pi  }{ T }  T =Time periodNow, at  x=2cm   v=|a|   \Rightarrow { \omega  }^{ 2 }	imes 2=\omega \sqrt { { 3 }^{ 2 }-{ 2 }^{ 2 } }  \Rightarrow { \omega  }=\cfrac { \sqrt { 5 }  }{ 2 }  \Rightarrow \cfrac { 2\pi  }{ T } =\cfrac { \sqrt { 5 }  }{ 2 }  \Rightarrow T=\cfrac { 4\pi  }{ \sqrt { 5 }  }

Question 5

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 = \dfrac{a}{2}  will be:

Accepted Answer

As simple pendulum performs simple harmonic motion. 	herefore velocity, v = \omega \sqrt {a^{2} - x^{2}} At,  x = \dfrac {a}{2} v = \dfrac {2\pi}{T} \sqrt {a^{2} -\left (\dfrac {a}{2}ight )^{2}} = \dfrac {2\pi}{T} \dfrac {\sqrt {3a^{2}}}{2} = \dfrac {\pi a\sqrt {3}}{T}

Question 6

The kinetic energy and potential energy of a particle executing SHM will be equal when displacement ( amplitude =a ) is

Accepted Answer

K.E.=P.E.\ \Rightarrow \frac { 1 }{ 2 } m{ \omega  }^{ 2 }({ a }^{ 2 }-{ x }^{ 2 })=\frac { 1 }{ 2 } m{ \omega  }^{ 2 }{ x }^{ 2 }\ \Rightarrow { a }^{ 2 }-{ x }^{ 2 }={ x }^{ 2 }\ \Rightarrow 2{ x }^{ 2 }={ a }^{ 2 }\ \Rightarrow x=a/\sqrt { 2 }

Question 7

A simple pendulum takes  32 s to complete  20  oscillations, what is the time period of the pendulum ?

Accepted Answer

Time taken to complete  20  oscillations  =32 sTime taken to complete  1  oscillation  =32/20 s  1.6 sTime period of a pendulum is time taken by it to complete  1  oscillation.Time period of pendulum is  1.6  seconds.

Question 8

Two masses  m_1  and  m_2  are suspended together by a massless spring of constant K as shown in figure. When the masses are in equilibrium,  m_1  is removed without disturbing the system. Find the angular frequency and amplitude of oscillation of  m_2 .

Accepted Answer

If only mass  { m }_{ 2 }  is present the extension of spring  l  is given by  { m }_{ 2 }g=kl\Rightarrow l=\cfrac { { m }_{ 2 }g }{ k }  , where  k =Spring constantNow with  { m }_{ 1 }  and  { m }_{ 2 }  the extension be  l'   ({ m }_{ 1 }+{ m }_{ 2 })g=kl' l'=\cfrac { { m }_{ 1 }+{ m }_{ 2 } }{ k }  When  { m }_{ 1 }  is removed,  (l'-1)  acts as amplitude of oscillation A=(l'-l)=(\cfrac { { m }_{ 1 }+{ m }_{ 2 } }{ k } )g-\cfrac { { m }_{ 2 }g }{ k }  \Rightarrow A=\cfrac { { m }_{ 1 }g }{ k }  Now, if  { m }_{ 2 }  is removed the angular frequency of oscillation  \omega =\sqrt { \cfrac { k }{ { m }_{ 2 } }  }  (This is same as the case of  { m }_{ 2 }  attach to spring) * N.B: When  { m }_{ 1 }  is removed the spring mass system will oscillates with the same angular frequency as the case of spring mass system with only mass  { m }_{ 2 }

Question 9

The displacement time graph of a particle executing S.H.M. is as shown in the figure. The corresponding force-time graph of the particle is:

Accepted Answer

Given displace-time graph indicates a SHM with sinusoidal variation.We know, for a SHM, force  F=-\omega^2y , where  y  (displacement) still varies sinusoidally with time.So,  \omega^2  being positive always, force-time graph will be a negative-sinusoid which is depicted in option (C).

Question 10

Two identical springs of constant are connected in series and parallel as shown in figure. A mass  m  is suspended from them. The ratio of their frequencies of vertical oscillations will be

Accepted Answer

We know that,  n=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} In series,  k=\dfrac{k_1k_2}{k_1+k_2}=\dfrac{k^2}{2} In parallel,  k=k_1+k_2=2k 	herefore, n_1=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{2m}} and  n_2 = \dfrac{1}{2\pi}\sqrt{\dfrac{2k}{m}} hence,  n_1:n_2=1:2