CLASS 12 PHYSICS MAHARASHTRA BOARD CHAPTER 5 SOLUTIONS
Q1. (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
(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 + m/4). Maximum kinetic energy of the body is
Q8. 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. [Ans: 4.33 cm]
Q9. In SI units, the differential equation of an S.H.M. d^2x—dt^2= —36x . Find its frequency and period. [Ans: 0.955 Hz, 1.05 s]
Q10. A needle of a sewing machine moves along a path of amplitude 4 cm with frequency 5 Hz. Find its acceleration (1/30) s after it has crossed the mean 30 position. [Ans: 34.2 m/s2]
Q11. Potential energy of a particle performing linear S.H.M is 0.1 pi^2 x^2 joule. If mass of the particle is 20 g, find the frequency of S.H.M. [Ans: 1.581 Hz]
Q12. 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. [Ans: 6.324 cm/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. [Ans: 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. [Ans: 1.48 x 10-2N]
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. [Ans: Decreases by 0.0051 m]
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? [Ans: 4 cm]
17. A particle performing linear S.H.M. of period 27π seconds about the mean position 0 is observed to have a speed of b SQRT(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. [Ans: π/3 s]
18. The period of oscillation of a body of mass m1 suspended from a light spring is T. Whcn 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 m, [Ans: 1/3]
Q19. The displacement of an oscillating particle is given by x = asincot +bcoscot where a, b and w are constants. Prove that the particle performs a linear S.H.M. with amplitude A=sq root (a^2 +b^2)
Q20. Two parallel S.H.M.s represented by xi = 5sin (47 pi t + pi/3) cm and x2 = 3sin (4 pi t + pi/4) cm are superposed on a particle. Determine the amplitude and epoch of the resultant S.H.M. [Ans: 7.936 cm, 54° 23']
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 conditions. [Ans: 0.04133 N m]
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/m^2. The magnet has moment of inertia 3 x 10^-6 kg/m2 and magnetic moment 3 A m^2. [Ans:38.19 osc/min.] ^
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? [Ans: nmax = 1/s, E/m = 1.25 J/kg]
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Below topics have been covered in this lecture: Conditions for Linear
SHM, Simple Harmonic Motion Key Terms, Mean Position, Amplitude in SHM,
Time Period and Frequency of SHM,
Phase in SHM, Phase Difference, Simple Harmonic Motion Equation and its
Solution,
Solutions of Differential Equations of SHM, Quantitative Analysis of
SHM, Velocity of a particle executing Simple Harmonic Motion, Kinetic
Energy of a Particle in SHM, SHM as a Projection of Circular Motion
x=displacement from mean positions, v=veleocity at position x,
a=acceleration at position x, ϕ- initial phase angle, w=angular
frequency, k=spring constant, A=amplitude
TIMESTAMP
00:00 overview (x,v,a, KE,PE, TE)
03:58 k-Spring Constant
05:06 equations overview
05:53 w,T,f discussion
11:52 how to study SHM
13:15 definition of SHM
18:47 derivation starts
23:45 Mean and Extreme values
27:45 v^2=w^2(a^2-x^2)
33:14 x=Asin(ωt+ϕ) calculations
35:00 ϕ discussion
47:14 v=Awcos(wt+ϕ) calculations
48:30 a=-Aw^2sin(ωt+ϕ) calculations
49:10 T calculations
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