LinkedIn Insight Solutions for Exercises of Oscillations - Grad Plus

Solutions for Exercises of Oscillations

1. Choose the correct option.

i) A particle performs linear S.H.M. starting from the mean position. Its amplitude is A and time period is T. At the instance when its speed is half the maximum speed, its displacement x is

$latex (A)\;\sqrt{\frac{3\;}2}\;\;A$

$latex (B)\;\frac2{\sqrt3}\;A$

$latex (C)\;\frac A2$

$latex (D)\;\frac1{\sqrt2}\;A$

ii) A body of mass 1 kg is performing linear S.H.M. Its displacement x (cm) at t (second) is given by x = 6 sin (100t + π/4). Maximum kinetic energy of the body is

(A) 36 J

(B) 9 J

(C) 27 J

(D) 18 J

iii) The length of second’s pendulum on the surface of earth is nearly 1 m. Its length on the surface of moon should be [Given: acceleration due to gravity (g) on moon is 1/6 th of that on the earth’s surface]

(A) 1/6 m

(B) 6 m

(C) 1/36 m

$latex \left(D\right)\;\frac1{\sqrt6}m$

iv) Two identical springs of constant k are connected, first in series and then in A metal block of mass m is suspended from their combination. The ratio of their frequencies of vertical oscillations will be in a ratio

(A) 1:4

(B) 1:2

(C) 2:1

(D) 4:1

v) The graph shows variation of displacement of a particle performing H.M. with time t. Which of the following statements is correct from the graph?

(A) The acceleration is maximum at time

(B) The force is maximum at time 3T/4.

(C) The velocity is zero at time T/2.

(D) The kinetic energy is equal to total energy at time T/4.


2. Answer in brief.

i)  Define linear simple harmonic motion.

ii) Using differential equation of linear S.H.M, obtain the expression for (a) velocity in S.H.M., (b) acceleration in H.M.

iii) Obtain the expression for the period of a simple pendulum performing S.H.M.

iv) State the laws of simple pendulum.

v) Prove that under certain conditions a magnet vibrating in uniform magnetic field performs angular S.H.M.


3. Obtain the expression for the period of a magnet vibrating in a uniform magnetic field and performing S.H.M.


4. Show that a linear S.H.M. is the projection of a U.C.M. along any of its diameter.


5. Draw graphs of displacement, velocity and acceleration against phase angle, for a particle performing linear S.H.M. from (a) the mean position (b) the positive extreme position. Deduce your conclusions from the graph.


6. Deduce the expressions for the kinetic energy and potential energy of a particle executing S.H.M. Hence obtain the expression for total energy of a particle performing S.H.M and show that the total energy is conserved. State the factors on which total energy depends.


7. Deduce the expression for period of simple pendulum. Hence state the factors on which its period depends.


8. At what distance from the mean position is the speed of a particle performing S.H.M. half its maximum speed. Given path length of S.H.M. = 10 cm.

$latex v\;=\frac12\;\;v_{max},\;x=?$

$latex We\;know\;that.\omega\sqrt{a^2-x^2}=v$

$latex \therefore\omega\sqrt{a^2-x^2}=\left(\frac12\right)v_{max}$

$latex But,\;v_{max}=\;a\omega,\;hence\;we\;get,$

$latex \therefore\;\omega\sqrt{a^2-x^2}\;=\;\left(\frac12a\omega\right)$

$latex \;\therefore\;2\sqrt{a^2-x^2}\;=\;a$

Squaring on both sides we get,

$latex \;4x^2\;=\;a^2$

$latex \therefore\;3a^2\;=\;4x^2\;$

$latex x^2\;=\;\frac{3\times\left(5^2\right)}4$

$latex x^2\;=\frac{75}4\;=\;18.75\;cm$

$latex \therefore x=4.33\;cm$


9. In SI units, the differential equation of an S.H.M. is  $latex \frac{d^2x}{dt^2}$=-36x . Find its frequency and period.

$latex \frac{d^2x}{dt^2}=-36x.\;n=?\;and\;T=?$

We know that, differential equation of a body performing S.H.M is given by,

$latex \frac{d^2x}{dt^2}=-\omega^2x$

Comparing given expression with this std expression,

$latex \omega^2=\;36\;\;\;\therefore\;\omega\;=\;6\;$

Now, we hvae,

$latex \omega\;=\;\frac{2\pi}T$

$latex \;\therefore\;T\;=\;\frac{2\pi}\omega$

$latex \therefore\;T\;=\;\frac{2\times3.14}6$

$latex \therefore\;T\;=\;\frac{3.14}3\;=\;1.05\;sec$

Now ω = 2πn

$latex \therefore\;n\;=\;\frac\omega{2\pi}$

$latex \therefore\;n\;=\;\frac6{6.28}$

$latex \therefore\;n\;=\;0.955\;Hz$


10. A needle of a sewing machine moves along a path of amplitude 4 cm with frequency 5 Hz. Find its acceleration($latex \frac1{30}$) s after it has crossed the mean position.

a = 4 cm = 4 × 10-2m, n = 5 Hz,

accln = ? after t =1/30 sec,

We know that,

$latex T=\frac{2\pi}\omega=\frac{2\pi}{2\pi n}$

$latex T\;=\;\frac{2\pi}\omega$

$latex T\;=\;\frac{2\pi}{2\pi n}\;=\;\frac15s$

Now, for S.H.M. we have,

$latex x\;=\;a\sin\omega t\;=\;a\sin\frac{2\pi}T\times t$

$latex =\;4\;\times\;10^{-2}\;\times\;\sin\;\left(10\pi\times\frac1{30}\right)$

$latex =\;4\;\times\;10^{-2}\;\times\;\sin\;\left(\frac\pi3\right)$

$latex=\;4\;\times\;10^{-2}\;\times\frac{\sqrt3}2$

$latex =\;2\sqrt3\;\times\;10^{-2}m$

$latex And\;accl^n\;=\;-\omega^2x\;$

$latex =\;100\pi^2\;\times\;2\sqrt{3\;}\times\;10^{-2}$

$latex =\;2\;\times\;3.14\;\times\;3.14\;\times\;\sqrt3\;$

$latex \therefore accl^n=34.15\;m/s^2$


11. Potential energy of a particle performing linear S.H.M is 0.1 π2 x2 If mass of the particle is 20 g, find the frequency of S.H.M.

P.E = 0.1πx Joule , m = 20g = 20 × 10-3 kg,

n = ?

We know that, potential energy for a particle performing S.H.M. is given as,

$latex P.E=\frac12m\omega^2x^2$

$latex \therefore\;\frac12m\omega^2x^{2\;}=\;0.1\pi^2x^2$

$latex \therefore\;m\omega^2x^2\;=\;0.2\pi^2x^2$

$latex \therefore\;\omega^{2\;}=\;\frac{0.2\pi^2}{20\times10^{-3}}$

$latex \therefore\;\left(2\pi n\right)^2\;=\;\frac{0.2\pi^2}{20\times10^{-3}}$

$latex \therefore\;4n^2\;=\;\frac{0.2\times10^3}{20}\\$

$latex \therefore\;4n^2\;=\;\frac{200}{20}\\$

$latex \therefore\;n^2\;=\;2.5$

$latex \therefore\;n\;=\;1.581Hz\;$


12. The total energy of a body of mass 2 kg performing S.H.M. is 40 J. Find its speed while crossing the centre of the path.

m = 2 kg ,T.E = 40 J, v = ?

We know that,

$latex T.E=\frac12m\omega^2a^2$

$latex \therefore\;40\;=\;\frac12m\omega^2a^2$

$latex \therefore\;80\;=\;2\;\times\;\omega^2a^2$

$latex \therefore\;40\;=\;\omega^2a^2$

$latex \therefore\;80\;=\;2\;\times\;\omega^2a^2$

$latex\therefore\;v\;=\;\sqrt{40}$

$latex \therefore\;v\;=\;2\sqrt{10}\;=\;2\;\times\;3.162$

$latex \therefore\;v_{max}\;=\;6.324\;m/s$


13. A simple pendulum performs S.H.M of period 4 seconds. How much time after crossing the mean position, will the displacement of the bob be one third of its amplitude.

T = 4 sec, x = 1/3a, t = ?

We know that,

i.e x = asinωt

$latex \therefore\;\frac13a\;=\;a\sin\omega t$

$latex \therefore\;\frac13\;=\;\sin\left(\frac{2\pi}T\times t\right)$

$latex \therefore\;\frac13\;=\;\sin\left(\frac\pi2\times t\right)$

$latex \therefore\;\sin^{-1}\left(\frac13\right)\;=\;\frac\pi2t$

$latex \therefore\;19.5^\circ\;=\;\frac\pi2t$

$latex \therefore\;19.5\;\times\;\frac\pi{180}=\frac\pi2t$

$latex \therefore\;t\;=\;0.2163\;s$


14. A simple pendulum of length 100 cm performs S.H.M. Find the restoring force acting on its bob of mass 50 g when the displacement from the mean position is 3 cm.

???? = 100 cm = 1m , F = ?

m = 50gm = 50 × 10-3 kg, x = a = 3 × 10-2 m

We know that,

$latex n=\frac1{2\pi}\sqrt{\frac gl}$

$latex\therefore\;n\;=\frac1{6.28}\sqrt{\frac{9.8}1}$

$latex \therefore\;n\;=\;\frac{3.14}{2\times3.14}\;=\;\frac12Hz$

$latex \therefore\;\omega\;=\;2\pi n$

$latex \therefore\;\omega\;=\;2\;\times\;\pi\;\times\frac12$

$latex \therefore\;\omega\;=\pi\;rad/s$

And, F = −kx

$latex \therefore\;F\;=\;-m\omega^2x\dots\dots Since,\omega^2\;=\frac km$

$latex \therefore\;F\;=\;m\omega^2x$

$latex \;\therefore\;F\;=\;50\;\times\;10^{-3}\;\times\;{(3.14)}^2\;\times\;3\;\times\;10^{-2}$

$latex \therefore\;F\;=\;150\;\times\;9.856\times\;10^{-5}$

$latex \therefore\;F\;=\;1.478\;\times\;10^{-2}N\;\;$


15. Find the change in length of a second’s pendulum, if the acceleration due to gravity at the place changes from 9.75 m/s2 to 9.8 m/s2.

Second’s Pendulum i.e. T = 2s,

g1 = 9.75m/s2, g2 = 9.8 m/s2, ∆l =?

Period of the pendulum is given as,

$latex T=2\pi\sqrt{\frac lg}$

$latex T=2\pi\sqrt{\frac lg}\;\;\;\;\;\;\;\therefore l=\frac g{\pi^2}$

$latex \therefore\;I_1\;=\;\frac{9.75}{\left(3.14\right)^2}\\$

$latex \;\therefore\;I_1\;=\;0.9878m,\\$

$latex \;I_2\;=\;\frac{9.8}{\left(3.14\right)^2}\\$

$latex \;\;I_2\;=\;0.9929m\\$

$latex \therefore\;\triangle l\;=\;0.9929\;-\;0.9878\\$

$latex \therefore\;\triangle l\;=\;0.0051m\\$


16. At what distance from the mean position is the kinetic energy of a particle performing S.H.M. of amplitude 8 cm, three times its potential energy?

x = ?, K.E = 3 P.E,

a = 8cm = 8 × 10-2m

We know that, for S.H.M.

$latex K.E\;=\;\frac12m\omega^2(a^2-x^2)\\$

$latex \&\;P.E=m\omega^2x^2\\$

Hence according to given data,

$latex \frac12m\omega^2(a^2-x^2)\;=\frac32m\omega^2x^2\\$

$latex m\omega^2a^2\;-\;m\omega^2x^2\\$

$latex =\;3m\omega^2x^2\\$

$latex \therefore\;m\omega^2a^2\;=\;4m\omega^2x^2\\$

$latex \therefore\;a^2\;=\;4x^2\\$

$latex \therefore\;x^2\;=\;\frac{a^2}4\\$

$latex \therefore\;\;x\;=\;\frac a2\\$

$latex \therefore\;x=\frac82\times10^{-2}\\$

$latex \therefore\;x\;=\;4\;\times\;10^{-2\;}=\;4cm\\$


17. A particle performing linear S.H.M. of period 27c seconds about the mean position 0 is observed to have a speed of b√3 m / s , when at a distance b (metre) from 0. If the particle is moving away from 0 at that instant, find the time required by the particle, to travel a further distance b.

$latex T\;=\;2\pi\;s,\;v\;=\;b\sqrt3m/s\;at\;x\;=\;b\\$

t = ? for further distance b.

We know that,

$latex v\;=\omega\sqrt{a^2-x^2}\\$

$latex \therefore\;b\sqrt{3\;}\;=\;\frac{2\pi}T\sqrt{a^2-b^2}$

$latex \therefore\;b\sqrt{3\;}\;=\;\frac{2\pi}{2\pi}\sqrt{a^2-b^2}$

$latex \therefore\;b\sqrt{3\;}\;=\;\sqrt{a^2-b^2}$

$latex \therefore\;3b^2\;=\;a^2-b^2$

$latex\therefore\;a^2\;=\;4b^2$

$latex \therefore\;a\;=\;2b$

Hence 2b is the extreme position, hence time required to travel 2b from mean position is $latex \frac T4\;i.e\;\frac\pi2s$

Time required to travel distance b from the mean position is given by, x = asinωt

$latex \therefore\;b\;=\;(2b)\sin\left(\frac{2\pi}Tt\right)$

$latex \therefore\;1\;=\;2\sin(t)$

$latex \therefore\;\sin t\;=\;\frac12\;\;\;\;\;\;\;i.e.\;t\;=\;\frac\pi6s$

Hence time required to travel further distance b from b is given as,

$latex t_{required}\;=\;\frac T4\;-\;t$

$latex t_{required}\;=\;\frac\pi2\;-\;\frac\pi6$

$latex t_{required}\;=\;\frac\pi3s$


18. The period of oscillation of a body of mass m1 suspended from a light spring is T .When a body of mass m2 is tied to the first body and the system is made to oscillate, the period is 2T. Compare the masses m1 and m2.

We know that,

$latex T=2\pi\sqrt{\frac km}$

Hence for rest of the conditions constant,

we have,$latex T\propto\sqrt m$

$latex \frac T{T’}\;=\;\frac{\sqrt m}{\sqrt{m’}}$

$latex \therefore\frac m{m’}\;=\;\left(\frac T{T’}\right)^2$

Given that, m = m1 & m’ = m1 + m2

also T’ = 2T

$latex \therefore\frac{m_1}{m_1+m_2}=\left(\frac T{2T}\right)^2=\frac14$

$latex \therefore\;4m_1\;=\;m_1\;+\;m_2$

$latex \therefore\;3m_1\;=\;m_2$

$latex \therefore\frac{m_1}{m_2}=\frac13$


19. The displacement of an oscillating particle is given by x= asinωt + bcosωt where a, b and ω are constants. Prove that the particle performs a linear S.H.M. with amplitude $latex A\;=\;\sqrt{a^2+b^2}$


20. Two parallel S.H.M.s represented by x1 = 5sin (4π t + π/3) cm and x2 = 3sin (4πt + π/4) cm are superposed on a Determine the amplitude and epoch of the resultant S.H.M.

$latex \therefore\;x_1\;=\;5\sin\left(4\pi t+\frac\pi3\right)cm,$

$latex \;x_2\;=\;3\sin\;\left(4\pi t+\frac\pi4\right)cm$

$latex R\;=\;?,\;\delta\;=\;?$

We know that,

$latex R\;=\;a_1^2+a_2^2+2a_1a_2\cos(\alpha_1-\alpha_2)$

From given data,

$latex a_1\;=\;5,\;a_{2\;}=\;3,$

$latex \alpha_1\;=\;\frac\pi3,\;\alpha_2\;=\;\frac\pi4$

$latex \therefore R=\sqrt{25+q+2\times5\times3\cdot\cos\left(\frac\pi3-\frac\pi4\right)}$

$latex \therefore\;R\;=\;\sqrt{35+30\times0.966}$

$latex \therefore\;R\;=\;\sqrt{34+3\times9.6}$

$latex \therefore R=\sqrt{62.8}=7.925\;cm$

Now we have,

$latex \therefore\delta=\tan^{-1}\left(\frac{a_1\sin\alpha_1+a_2\sin\alpha_2}{a_1\cos\alpha_1-a_2\cos\alpha_2}\right)$

$latex \therefore\delta=\tan^{-1}\left(\frac{5\sin{\displaystyle\frac\pi3}+3\sin{\displaystyle\frac\pi4}}{5\cos{\displaystyle\frac\pi3}+3\cos{\displaystyle\frac\pi4}}\right)$

$latex \therefore\delta=\tan^{-1}\left(\frac{5{\displaystyle\frac{\sqrt3}2}+3\times{\displaystyle\frac1{\sqrt2}}}{5\times{\displaystyle\frac12}+3\times{\displaystyle\frac1{\sqrt2}}}\right)$

$latex \therefore\;\delta\;=\;\tan^{-1}\left(\frac{5\sqrt3+3\sqrt2}{5+3\sqrt2}\right)$

$latex \tan^{-1}\;=\;\frac{5\times1.732+3\times1.414}{5+4.242}$

$latex \therefore\;\delta\;=\;\tan^{-1}\left(\frac{12.90}{9.242}\right)$

$latex \therefore\;\delta\;=\;\tan^{-1}(1.39)\;=\;54^\circ23’$


21. A 20 cm wide thin circular disc of mass 200 g is suspended to a rigid support from a thin metallic string. By holding the rim of the disc, the string is twisted through 60° and released. It now performs angular oscillations of period 1 second. Calculate the maximum restoring torque generated in the string under undamped (π3 ≈ 31)

r = 10cm = 10 × 10-2m =10-1m

m = 200g = 200 × 10-3kg ,

$latex \theta=60^\circ=\frac\pi6,\;T\;=\;1s,\;\tau\;=?\;$

We know that,

$latex T\;=\;\frac{2\pi}\omega$

$latex \therefore\;\omega\;=\;\frac{2\pi}T\;=\;2\pi\;rad/s$

The restoring torque is given as, τ = Iα

Here I is the moment of inertia and angular acceleration.

$latex\therefore\;\tau\;=\;\left(\frac{mr^2}2\right)\left(\omega^2\theta\right)$

$latex \therefore\;\tau\;=\;\left[\frac{200\times10^{-3}\times\left(10^{-1}\right)^2}2\right]\left[\left(2\pi\right)^2\times\frac\pi6\right]$

$latex\therefore\;\tau\;=\;\frac{\pi^3}{1500}\;=\;\frac{31}{1500}$

$latex \therefore\;\tau\;=\;0.0267Nm$


22. Find the number of oscillations performed per minute by a magnet is vibrating in the plane of a uniform field of 1.6 x 10-5 Wb/m2. The magnet has moment of inertia 3 x 10-6 kg/m2 and magnetic moment 3 A m2.

B = 1.6 × 10-5Wb/m , I = 3 × 10-6kgm2

μ = 3Am , Number of oscillations per minute = ?

For a magnet is vibrating in the plane of a uniform field,

$latex We\;have,\;T=2\pi\sqrt{\frac1{\mu B}}$

Oscillations per second,

$latex =n=\frac1t=\frac1{2\pi}\sqrt{\frac{\mu B}1}$

$latex \therefore\;n\;=\;\frac1{2\pi}\sqrt{\frac{3\times\left(1.6\times10^{-5}\right)}{3\times10^{-6}}}$

$latex \therefore\;n\;=\;\frac1{2\pi}\sqrt{\frac{\left(16\times10^{-6}\right)}{10^{-6}}}$

$latex i.e.\;n\;=\;\frac1{2\pi}\sqrt{16}$

$latex \therefore n=\frac2\pi\;oscillations\;per\;second$

∴ Number of oscillations per minute,$latex =n\times60=\frac{120}\pi$

∴ Number of oscillations per minute=38.19


23. A wooden block of mass m is kept on a piston that can perform vertical vibrations of adjustable frequency and amplitude. During vibrations, we don’t want the block to leave the contact with the piston. How much maximum frequency is possible if the amplitude of vibrations is restricted to 25 cm? In this case, how much is the energy per unit mass of the block? (g ≈ π2 ≈ 10 m s -2 )

During vibrations, we don’t want the block to leave the contact with the piston. That means at the extreme (top) position, acceleration must be less than or equal to the acceleration due to gravity i.e. g.

For SHM we have,

$latex Accleration\;=\;\omega^2x$

Hence at extreme position, Accln = ω2a

Given that,

Amplitude, a = 25cm = 25 × 10-2m

∴ g = ω2 × 25 × 10-2

$latex \therefore\;\omega^2=\frac{10}{25\times10^{-2}}=\frac{10^2}{25}$

$latex \therefore\;\omega\;=\;\frac{10\sqrt{10}}5$

$latex \therefore\;n\;=\;\frac{10\times\pi}{5\times2\times\pi}\;as\;\pi^2=10$

∴ n = 1Hz

We know that, energy in SHM is ,

$latex E\;=\;\frac12m\omega^2a^2$

Hence energy per unit mass is given as,

$latex \frac Em\;=\;\frac12\omega^2a^2$

$latex \frac Em\;=\;\frac12\times\left(\frac{10\sqrt{10}}5\right)^2\times\left(25\times10^{-2}\right)^2$

$latex \frac Em\;=\;1.25$

$latex \therefore\;\frac Em\;=\;1.25\;J/kg$

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