![1.5 : Superposition of Two Perpendicular Harmonic
Oscillations Having Equal Frequencies
(a) Analytical method :
Let x = asin(𝜔t + a) --------11
and y= b sin 𝜔 t --------------12
be the displacements produced by two SHM's acting
simultaneously on a particle in perpendicular directions i.e. one
along X-axis and another along Y-axis.
Sin𝜔t =
𝑦
𝑏
, -----------from equation 11
Cos 𝜔t=√1 −
𝑦2
𝑏2, using identity sin2
𝜔t + 𝑐𝑜𝑠2
𝜔t = 1,
𝜘
𝑎
= 𝑠𝑖𝑛(𝜔t + 𝛼) = 𝑆𝑖𝑛 (𝐴 + 𝐵) = sin𝜔t. cos𝛼 + cos𝜔t. sin𝛼
=[
𝑦
𝑏
𝑐𝑜𝑠 𝛼 + √1 −
𝑦2
𝑏2 . 𝑠𝑖𝑛𝛼 ]
𝜘
𝑎
−
𝑦
𝑏
𝑐𝑜𝑠𝛼 = √1 −
𝑦2
𝑏2
. 𝑠𝑖𝑛𝛼
Squaring this equation we get
𝜘2
𝑎2
+
𝑦2
𝑏2
𝑐𝑜𝑠𝛼
2
−
2𝜘𝑦
𝑎𝑏
𝑐𝑜𝑠 𝛼 = (1 −
𝑦2
𝑏2
). 𝑠𝑖𝑛𝛼
2](https://image.slidesharecdn.com/superpositionofharmonicoscillator-2-230221110311-96b9ef97/85/Superposition-of-Harmonic-oscillator-2-docx-1-320.jpg)
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- 1. 1.5 : Superposition of Two Perpendicular Harmonic Oscillations Having Equal Frequencies (a) Analytical method : Let x = asin(𝜔t + a) --------11 and y= b sin 𝜔 t --------------12 be the displacements produced by two SHM's acting simultaneously on a particle in perpendicular directions i.e. one along X-axis and another along Y-axis. Sin𝜔t = 𝑦 𝑏 , -----------from equation 11 Cos 𝜔t=√1 − 𝑦2 𝑏2, using identity sin2 𝜔t + 𝑐𝑜𝑠2 𝜔t = 1, 𝜘 𝑎 = 𝑠𝑖𝑛(𝜔t + 𝛼) = 𝑆𝑖𝑛 (𝐴 + 𝐵) = sin𝜔t. cos𝛼 + cos𝜔t. sin𝛼 =[ 𝑦 𝑏 𝑐𝑜𝑠 𝛼 + √1 − 𝑦2 𝑏2 . 𝑠𝑖𝑛𝛼 ] 𝜘 𝑎 − 𝑦 𝑏 𝑐𝑜𝑠𝛼 = √1 − 𝑦2 𝑏2 . 𝑠𝑖𝑛𝛼 Squaring this equation we get 𝜘2 𝑎2 + 𝑦2 𝑏2 𝑐𝑜𝑠𝛼 2 − 2𝜘𝑦 𝑎𝑏 𝑐𝑜𝑠 𝛼 = (1 − 𝑦2 𝑏2 ). 𝑠𝑖𝑛𝛼 2
- 2. 𝜘2 𝑎2 + 𝑦2 𝑏2 𝑐𝑜𝑠𝛼 2 + 𝑦2 𝑏2 𝑠𝑖𝑛𝛼 2 − 2𝜘𝑦 𝑎𝑏 𝑐𝑜𝑠 𝛼 = 𝑠𝑖𝑛𝛼 2 𝜘2 𝑎2 + 𝑦2 𝑏2 (𝑐𝑜𝑠𝛼 2 + 𝑠𝑖𝑛𝛼 2 ) − 2𝜘𝑦 𝑎𝑏 𝑐𝑜𝑠 𝛼 = 𝑠𝑖𝑛𝛼 2 𝜘2 𝑎2 + 𝑦2 𝑏2 − 2𝜘𝑦 𝑎𝑏 𝑐𝑜𝑠 𝛼 = 𝑠𝑖𝑛𝛼 2 ---------------13 Equation 13 is general equation for ta ellipse which represents the locus of the particle subjected two SHM’s at right angles to each other simultaneously Depending upon the phase difference (𝛼),we have different shapes of the locus of the particle called Lissajous figures. i)if 𝛼 = 0𝑜𝑟 2𝜋, then sin 𝛼 = 0 and cos𝛼=1 from equation 13 we get 𝜘2 𝑎2 + 𝑦2 𝑏2 − 2𝜘𝑦 𝑎𝑏 = 0 ( 𝜘 𝑎 − 𝑦 𝑏 )2 =0taking roots of both sides we get 𝜘 𝑎 = 𝑦 𝑏 and y = 𝑏 𝑎 ϰ which is the equation of straight line DB in following fig.
- 3. ii)If a = 𝜋,then sin 𝞌=0 and cosϰ=-1 from equation 13 we get 𝜘2 𝑎2 + 𝑦2 𝑏2 + 2𝜘𝑦 𝑎𝑏 = 0 Then y = − 𝑏 𝑎 ϰ represent the straight line (AC)in above fig. iii)If a = 𝜋 2 or 3𝜋 2 . then sin 𝛼 = 1 and cos 𝛼 = 0 equation 13 becomes 𝜘2 𝑎2 + 𝑦2 𝑏2 = 1, this represent an ellipse EHGF with semi-major axis (a) and semi-major axis (b)as shown in above fig. iv))If𝛼 = 𝜋 2 or 3𝜋 2 and a=b , then equation 13 becomes 𝜘2 + 𝑦2 = 𝑎2 , which represent a circlewith radius a. v) If 𝛼 = 𝜋 4 or 7𝜋 4 , then vibration is along an oblique ellipse KLMN shown in fig. 1.3 (i)
- 4. Vi) 𝛼 = 3𝜋 4 or 5𝜋 4 then resultant vibration is along an oblique ipse KLMN as shown in fig. 1.3 (ii). (ii) Graphical method : Consider a particle 0 subjected to two simple harmonic motions with same frequency (𝜔) but different amplitudes with a phase difference (𝛼). So the two SHM's are represented by, 𝞌 = a sin(𝜔t + a) and y = bsin 𝜔t Let 𝛼= 𝜋 2 be the phase difference. Acting in right angle with each other perpendicular.
- 5. Fig. 1.4 : Resultant vibration of two SHM's with same frequency (𝜔) but different amplitudes and phase difference 𝛼 = 𝜋 2 . Draw reference circles with centers𝐶1 and 𝐶2and radii a and b respectively as shown in fig. 1.4. Divide each circle into eight equal parts and mark as 0, 1, 2,--- 7, 8. The marks on the reference circle are such that they show a phase difference of 𝜋 2 .
- 6. When the particle 0 is subjected to simple harmonic motion along X-axis only, then the particle will vibrate along XX'. Similarly if 0 is subjected to SHM along Y-axis only, then the particle will vibrate along YY'. But when 0 is subjected to both the vibrations simultaneously, the resultant vibration will be an ellipse ABCDEFGHA as shown in fig. 1.4. iperposition of Harmonic Oscillations 13 At, t = 0 the positions on the reference circle are marked as '0', showing a phase difference of 𝜋 2 . Therefore, corresponding position of particle '0' will be A. At time t = 𝑇 8 (where T is period of vibration), the positions on the reference circle are marked as I'. Therefore,the resultant position of the particle-0 will be at B etc. as shown in fig. 1.4. Thus, resultant motion ofthe particle 0 will be along a reference ellipse ABCDEFGHA with the same angular frequency (co) or same period(T) as for two constituent SHM's. Thus at any instant of time the resultant motion is given by,
- 7. 𝜘2 𝑎2 + 𝑦2 𝑏2 = 1i.e. an ellipse But if a = b, then resultant motion is given by a the equation, 𝜘2 + 𝑦2 = 𝑎2 where a is the radius of the circle with 0 as center. Similarly, for different phase differences we can graphically find the resultant motion of the particle 0. 1.6 : Superposition of Two SHM's at Right Angles to each other and having frequencies in the ratio 2 : 1 (i) Analytical method : Let us consider two SHM's, one along X- axis and another along Y-axis with amplitudes a and b respeLtively and differing in phase by 900 . Also frequencies are in the ratio 2 : 1. So the analytical equations for two SHM's are x=asin(2𝜔t+ 𝛼) and y=bsin(𝜔t) .∴ 𝑦 𝑏 = sin 𝜔𝑡 and cos 𝜔𝑡 = √1 − 𝑦2 𝑏2 ∴ 𝜘 𝑎 = 𝑠𝑖𝑛(2𝜔𝑡 + 𝛼)
- 8. =sin2𝜔𝑡. 𝑐𝑜𝑠𝛼 + 𝑐𝑜𝑠2𝜔𝑡. 𝑠𝑖𝑛 𝛼 sin2A=2sinAcosA 𝜘 𝑎 =2sin 𝜔𝑡. 𝑐𝑜𝑠𝜔𝑡. 𝑐𝑜𝑠𝛼 + (1 − 2𝑠𝑖𝑛2 𝜔𝑡) 𝑠𝑖𝑛𝛼 =( cos 2 𝜔𝑡 = 𝑐𝑜𝑠2 𝜔𝑡 − 𝑠𝑖𝑛2 𝜔𝑡 = 1 − 𝑠𝑖𝑛2 𝜔𝑡 − 𝑠𝑖𝑛2 𝜔𝑡 ) cos2A=𝑐𝑜𝑠2 𝐴 − 𝑠𝑖𝑛2 𝐴 = 1 − 2𝑠𝑖𝑛2 𝜔𝑡 Now substituting for sin𝜔𝑡 and cos𝜔𝑡
- 12. Draw two reference circles with centres as C1 and C2 and radii a and b respectively. Then divide circle with centre C1 into 4- equal parts and the circle with centre C2 into 8-equal parts and mark them by 0. 1. 2,….8.
- 13. 1. At time t = 0, the positions on the reference circle are marked by 0. Hence the resultant position of the particle subjected to both the vibrations simultaneously will be A. 2. Similarly at time t = T/8, the positions on the reference circle correspond to '1' and hence the resultant position of the particle will be at B etc. Thus the resultant vibration of the particle subjected to two SHM's will be as shown in fig. 1.6 i.e. looks like figure 8 (eight). So, for different phase differences we can graphically find the shape of the curve along which the particle moves. 1.7 : Uses of Lissajous Figures (a) To determine the ratio of time periods or frequencies : From the observation of Lissajous figures, the number of times the curve 'ouches the horizontal and vertical sides of a rectangle bounding the Lissajous figure is found. If the curve touches horizontal side in times and vertical side n times, then
- 14. the ratio of time periods is n : n or the ratio of frequencies is n : tn. eg. in the fig. (1.6), the curve touches the horizontal side once and vertical side twice. Hence the ratio of the periods is 1 : 2 (or ratio of frequencies is 2 : 1). The ratio of frequencies is determined when the ratio is represented by simple integral numbers. Also if one of the frequency is known other can be calculated. (b) When frequencies are nearly equal : The unknown frequency of a tuning fork can be determined, if the frequency of other tuning fork is known. Let a tuning fork A with unknown frequency (n 1) and a tuning fork B with a known frequency (n) are sounded together to produce Lissajous figures When both sources start vibrating in phase, the resultant will be a straight line. As the time passes, the phase difference goes on increasing and hence the shape of resultant curve goes on changing i.e. different Lissajous figures are traced. When one of the sources has completed one vibration more than the other, the two vibrations will be in-phase and same Lissajous figure (i.e. straight line) will appear. This cycle of changes goes on repeating. If t is the time for one such cycle, then difference in frequency = p = 1 𝑡 ∴n1~ 𝑛 = 𝑝 ∴ 𝑛1 = 𝑛 ± 𝑝 … ..
- 15. Now fix little wax on the prong of tuning fork A and repeat the observation and find the difference in frequency as, P’= 1 𝑡′ 1)If t’ < t, then p'> p is the observation and if n1> n, then n1- n= p and n1’ — n =p'. For a loaded tuning fork frequency decreases n1' < n1 . ∴p' < p which contradicts the observation ( p' > p ). ∴n1 - n = p is wrong. Hence n > n1 ∴n – n1 =p or n1= n - p is the unknown frequency. (ii) If t'> t, then p' < p is the observation and if n1 > n , then n1- n = p . And n1’- n = p' for loaded A. Since n1 ' < n1 we see that p' < p which justifies the observation. ∴n1 = n+ p or n1= n+ p is unknown frequency of (A).