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Oscillations

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Published in: AIEEE | IIT JEE Mains
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Notes On Oscillations

Akhilesh K / Lucknow

4 years of teaching experience

Qualification: M.Sc (NIT Rourkela - 2019)

Teaches: All Subjects, English, Mathematics, Science, Chemistry, Physics, Algebra, IIT JEE Mains, AIPMT, NEET

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  1. OSCILLATIONS Periodic Motion and Oscillatory motion If a body repeats its motion along a certain path, about a fixed point, at a definite interval of time, it is said to have a periodic motion If a body moves to and fro, back and forth, or up and down about a fixed point in a definite interval of time, such motion is called an oscillatory motion. The body performing such motion is called an oscillator. Simple Harmonic Motion The periodic motion of a body about a fixed point, on a linear path, under the influence of the force acting towards the fixed point and proportional to displacement of the body from the fixed point is called a simple harmonic motion (SHM) A body performing simple harmonic motion is known as Simple Harmonic Oscillator (SHO) Simple harmonic motion is a special type of periodic motion in which The particle oscillates on a straight line. The acceleration of the particle is always directed towards a fixed point on the straight line. (iii) The magnitude of acceleration is proportional to the displacement of the particle from fixed point. x This fixed point is called the centre of the oscillation or mean position. 'Taking this point as origin O The maximum displacement of oscillator on either side of the mean position is called amplitude denoted by A The time required to complete one oscillation is known as periodic time (T) of oscillator In other words, the least time interval of time after which the periodic motion of an oscillator repeat itself is called a periodic time of the oscillator. Distance traveled by oscillator is 4A in periodic time The number of oscillations completed by simple harmonic oscillator in one second is defined as frequency. Sl unit is s-1 or hertz (H) It is denoted by f and f = I/T 2n times the frequency of an oscillator is called the angular frequency of the oscillator It is denoted by (D. Its Sl unit is rad s-1, (D = 21Tf. If we draw the graph of displacement of SHO against time as shown in figure, which is as shown in figure
  2. c: a. x o 4 4 2 Mathematical equation is the function of time x(t) = Asin(wt ++) eq(l) Here X(t) represents displacement at time t A = Amplitude (wt + 4)) = Phase = Initial phase (epoch) for a given graph O. If oscillation have started from negative x end (M) then initial phase is —n/ 2 If oscillation stated from positive end (N) then initial phase is Tt/2 w = angular frequency Velocity Velocity of oscillator is v(t) = wAcos(wt + (P ) v(t) — v(t) From eq(l) dx(t) v(t) = dt 1 — sin2 (Ot + Q)) A2 — + v(t) = ± O) ( A2 —X 2) Velocity is maximum at x = O or equilibrium position v = Velocity is minimum at x = A or extreme positions v= O Note that velocity is out of phase of displacement by TT/2 Acceleration dv(t) a(t) = dt a(t) = -02Asin(wt + $) a(t) = -02 X(t) At x = O ,acceleration is a = O And maximum at x = -I—A , a = c02A Note velocity is out of phase of displacement by IT
  3. Relation between simple harmonic motion and uniform circular motion In figure shown Q is a point moving on a circle of radius A with constant angular speed G) (in rad/sec). P is the perpendicular projection of Q on the horizontal diameter, along the x-axis. x P o x P cot+ o C02 A x Pax O x Let us take Q as the reference point and the circle on which it moves the reference circle. As the reference point revolves, the projected point P moves back and forth along the horizontal diameter Let the angle between the radius OQ and the x-axis at the time t=O be called 4. At any time t later the angle between OQ and the x-axis is (wt+ 4), the point Q moving with constant angular speed w. The x- coordinate of Q at any time is, therefore X = Acos(wt+ 4) i.e. P moves with simple harmonic motion Thus, when a particle moves with uniform circular motion, its projection on a diameter moves with simple harmonic motion. The angular frequency CD of simple harmonic motion is the same as the angular speed of the reference point. The velocity of Q is v = (DA. The component of v along the x-axis is = -vsin(wt+ $) - (DAsin(wt+ 4)), Which is also the velocity of p. The acceleration of Q is centripetal and has a magnitude, a = 02 A The component of 'a' along the x-axis is Ax = -acos(wt+ 4), Ax = -02 Acos(wt+ 4), Which is the acceleration of P The force law for simple harmonic motion We know that As a = F = -m02x(t) This force is restoring force According to Hook's law, the restoring force is given by F = -kx(t) With k as spring constant Thus k = m 02 angular frequency k m frequency of oscillation 1 27T k m
  4. Examples of simple harmonic motion Simple pendulum A simple pendulum consists of a heavy particle suspended from a fixed support through a light, inextensible and torsion less string The time period of simple pendulum can be found by force or torque method and also by energy method (a)Force method: the mean position or the equilibrium position of the simple pendulum is when 0 = O as shown in figure(i). the length of the string is l, and mass of the bob is m When the bob is displaced through distance the forces acting on it are shown in figure(ii) The restoring force acting on the bob to bring it to the mean position is F = - mgsine ( -ve sign indicates that force is directed towards the mean position) For small angular displacement x mgsinO Sino = x/l mgcos0 x F = —mg x mg (ii) Comparing it with equation of simple harmonic motion a 2 Time period 27T = 27T (b)Torque method: Now taking moment of force acting on the bob about O T = —(mgsin0)l Also from Newton's second law T = la From equation(l) and (2) — eq(l) Since 0 is small But Moment of inertia I = m12 — eq(2) la = —(mgsin0)l la —
  5. m12a (mg0)l (90 Comparing with simple harmonic motion equation = - 1 27T = 27T Note: component of force mgcos0 cannot produce torque because it passes through fixed point (c) Energy method: Let the potential energy at the mean position be zero. Let the bob is displaced through an angle '0'. Let its velocity be Then potential energy at the new position U = mgl(1-cos0) Kinetic energy at this instant K = (1/2) mv2 x (iii) Total mechanical energy at this instant E=U+K E = mgl(1-cos9) + (1/2) mv2 We know, in simple harmonic motion E dt mgl sinogg + mv dv dt dt Bur v = (DI dt d0 dv . mgl sine— + ml— dt dt dt dv g[sin0] + = constant dv dt dt g [Sino] But 0 = x/l
  6. x 1 But a=-G) x 2 Periodic time 1 27T 27T (a)SimpIe pendulum in a lift: If the simple pendulum is oscillating in a lift moving with acceleration a, then the effective g of the pendulum is geff = g±a + sign is taken when lift is moving upward ve sign is taken when lift is moving downward Hence periodic time of pendulum 1 (b)Simple pendulum in the compartment of a train If the simple pendulum is oscillating in a compartment of a train accelerating or retarding horizontally at the rate 'a' then the effective value of g is geff Hence periodic time of pendulum (c)Seconds pendulum The pendulum having the time-period of two seconds, is called the second pendulum. It takes one second to go from one end to the other end during oscillation. It also crosses the mean position at every one second Combinations of springs Series combination As show in figure consider a series combination of two massless spring of spring constant and
  7. where F is the restoring force. We know that F = -kx x =-F/k From the above equations, Time period T Frequency f In this system when the combination of two springs is displaced to a distance y, it produces extension Xl and in two springs of force constants kl and 1
  8. mg k 2 Frequency vertical plane after pulling mass m In this situation when a body is pulled lower through small displacement y. lower spring gets compressed by y, while upper spring elongate by y. hence restoring forces Fl and F2 set up in both these springs will act in the same direction. Net restoring force will be F kly -k2Y F= - (kl + k2) y If k' is the is equivalent spring constant then F = -k'y thus k' = kl + Now periodic time T T = 27T 1 27T k' m 1 27T If kl = k2 = k then 2k (ii) Two massless springs of equal lengths having force constants kl and k2 respectively are suspended vertically from a rigid support as shown in figure. At their free ends, a block of mass m having non-uniform density distribution is suspended so that spring undergoes equal extension In this situation two bodies are pulled down through a small distance y and the system is made to perform SHM in vertical plane Here, the springs have different force constants. Moreover the increase in their length is same. Therefore, the load is distributed equally between the springs. Hence, the restoring force developed in each spring is different. If Fl and h are the restoring forces set up due to extension of springs, then Fl = -kl y and h = k2Y Also the total restoring force F Fl + F2 F = - kly -1
  9. Differential equation of simple harmonic motion According to Newton's second law of motion F = ma Comparing this with F = -ky(t) d2y(t) dt2 d2y(t) dt2 d2y(t) dt2 d2y(t) d2y(t) m dt2 ky(t) k 02y(t) + 02 y (t) = O dt2 This is the second order differential equation of the simple harmonic motion. The solution of this equation is of the type Y(t) = A sinot or y(t)=Bcos(Dt Or any linear combination of sine and cosine function Y(t) = A sinot + Bcosot Total Mechanical Energy in Simple Harmonic Oscillator The total energy (E) of an oscillating particle is equal to the sum of its kinetic energy and potential energy if conservative force acts on it. Kinetic energy Kinetic energy of the particle of mass m is 1 1 2 —m02(A2 — x2) K = —mv 2 Potential energy — eq(l) From definition of SHM F = —kx the work done by the force during the small displacement dx is dW = -F.dx = -(-kx) dx = kx dx Total work done for the displacement x is, W = dw = k=02m x kx dx 1 mc02x dx 2 This work done is stored in the body as potential energy 1 U = —m02x2 (2)
  10. Total energy E = K + U Special cases E — —m02(A2 — x2) + E = —m02A2 (i) When the particle is at the mean position x= 0, from eqn (1) it is known that kinetic energy is maximum total energy is wholly kinetic Km ax = -mø2A2 and from eqn. (2) it is known that potential energy is zero. (ii) When the particle is at the extreme position y = +a, from eqn. (1) it is known that kinetic energy is zero and from eqn. (2) it is known that Potential energy is maximum. Hence the total energy is wholly potential. Um ax (iii)When y = A/2 nt(D2A2 K = —m02 A2 31 —mø2A2 —m02A2 If the displacement is half of the amplitude K and U are in the ratio 3 : 1,
  11. Angular Simple Harmonic Motion A body to rotate about a given axis can make angular oscillations. For example, a wooden stick nailed to a wall can oscillate about its mean position in the vertical plane 'The conditions for an angular oscillation to be angular harmonic motion are (i)When a body is displaced through an angle from the mean position, the resultant torque is proportional to the angle displaced (ii)This torque is restoring in nature and it tries to bring the body towards the mean position If the angular displacement of the body at an instant is 0 , then resultant torque on the body t = -k0 if the momentum of inertia is I, the angular acceleration is k 9 20 ——— eq(l) d20 dt2 Here Or k Solution of equation (1) is 0 = 00 sin(ot + Q) Where is the maximum angular displacement on either side. Angular velocity at time 't' s given by Physical pendulum An rigid body suspended from a fixed support constitutes a physical pendulum. o. c mg o 6 c mg As shown in figure is a physical pendulum. A rigid body is suspended through a hole at O. When the centre of mass C is vertically below O at a distance of the body may remain at rest. The body is rotated through an angle e about a horizontal axis OA passing through O and perpendicular to the plane of motion The torque of the forces acting on the body, about the axis OA is T = mglsine, here I = oc If momentum of inertia of the body about OA is l, the angular acceleration becomes
  12. For small angular displacement sine = 0 (mgl a Comparing with = - 029 27T = 27T mg I Torsional pendulum In torsional pendulum, an extended body is suspended by a light thread or wire. The body is rotated through an angle about the wire as the axis of rotation. The wire remains vertical during this motion but a twist '0' is produced in the wire. The twisted wire exerts a restoring torque on the body, which is proportional to the angle of the twist. oc -0 ; T = - ; k is proportionality constant and is called torsional constant of the wire. If I be the moment of inertia of the body about vertical axis, the angular acceleration is —k —9 = —029 k 21T Time period T — = 27T Two Body System In a two body oscillations, such as shown in the figure, a spring connects two objects, each of which is free to move. When the objects are displaced and released, they both oscillate. The relative separation Xl- gives the length of the spring at any time. Suppose its unscratched length is L; then x = (Xl — — L is the change in length of the spring, and F = kx is the magnitude of the force exerted on each particle by the spring as shown in figure. Applying Newton's second law separately to the two particles, taking force component along the x-axis, we get d2X1 d2X2 dt2 = —kx and ml dt2 Multiplying the first of these equations by and the second by ml and then subtracting, d2X1 ml dt2 d d 122 = —m2kx — mikx
  13. Since L is constant Equation (i) becomes Thus periodic time Damped oscillation This can be written as ml 1712 d2 — x2) = —kx (ml + 1112) dt2 d2 kx — — — eq(l) Here VI is known as reduced mass and has dimension of mass = —kx T = 27T — xo — d2x dt2 nil (ml + 1712) d2 dx dt k —x k Experimental studies showed that the resistive force acting on the oscillator in a fluid medium depends upon the velocity of the oscillator Thus resistive force or damping force acting on the oscillator is Fd = -nv Here b is damping constant and has Sl units kg/second. The negative sign indicates that the force Fd opposes the motion Thus, a damped oscillator oscillate under the influence of the following forces (i)Restoring force Fx = - kx and (ii) Resistive force Fd = -bv Net force F = FK + Fd According to second law of motion ma = -ky— bv d2x dt2 —ky — b dx dt
  14. —bt/ x(t) = A e 2m sin(dt + p) 2m is the amplitude of the damped oscillation and decreases exponentially Here A e with time The angular frequency G)' of the damped oscillator is given by b2 k m 4m2 Energy of oscillator 1 E = —kAe m Above equation is valid only if b
  15. + b202]1/2 [m2 (o; ox —1 cp = tan Vo Here m is the mass of oscillator, vo is velocity of oscillator, x is the displacement of oscillator. (i) for small damping factor m(coä — 02) >> bw Equation for amplitude becomes — 02)2 (ii) For large damping factor bo >> m(wå — 02) If when value of w approaches oothe amplitude becomes maximum. This phenomenon is known as resonance. The value of w for which resonance occurs is known as the resonant frequency.

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