Worked out Exaples. The potential energy function for the force between two atos in a diatoic olecules can be expressed as follows: a U(x) = b x / x6 where a and b are positive constants and x is the distance between atos. (i) At what value of x is U(x) equal to zero? (ii) At what values of x is U(x) iniu? (ii) Derive the expression for the force between two atos. Solution: U(x) = a b. x x 6 (i) For U(x) = 0 a b x x6 = 0. Or, ( a x 6 x6 b) = 0. Which s possible only when = 0 or ( a x 6 x6 b) = 0. Finally we get x = or x = ( a b )/6. (ii) For iniu value of U(x), du dx = 0. d ( a b dx x x6) = 0 Or, a 6b x + x 7 Or, a x7 (6b ) = 0 x = or, x = x 6 (a b )/6 (iii) Force F between two atos is given by F = du a 6b a 6b = ( dx x + x7) = x x 7. A particle oves in a potential energy field given by V = V 0 Ax + Bx.(i) Find an expression for the force acting on the particle (ii) Calculate force constant (iii) At what point does the force vanish? Is this point a stable equilibriu? Solution: (i) F = du = 0 ( A + Bx) = A Bx. dx (ii)if k is force constant, then F = kx, coparing it with F = A Bx, we get k = B. (iii) F = A Bx = 0 x = A. For the stable equilibriu, the potential B energy will be iniu i.e. here V 0 which occurs at the point x = A. B. If the displaceent of a oving particle at any ties is given by x = a cos ωt + b sin ωt. Show that otion is siple haronic. If a=, b=4, ω=. Find the period, axiu velocity and axiu acceleration. Hint: calculate dx dt and then d x dt. Here d x dt = ω (a cos ωt + b sin ωt) = ω x which proves the otion is siple haronic. Now A = a + b = + 4 =
5 c. Tie Period T= π =.4 =.4 sec. The axiu velocity v ω ax = ωa = 5 = 0 c/sec and axiu acceleration a ax = ω A = 5 = 0 c. sec 4. The equation of otion of a particle is x = sin( πt )c. Fine the period and axiu velocity of the particle. Hint: copare given equation x = sin( πt + π 4 + π 4 ) with general equation of siple haronic otion x = a sin(ωt + φ). Then a = c, ω = π rad, T = π = sec ω 4 sec. Maxiu velocity of the particle v ax = ωa = π c. sec 5. The equation of otion of an oscillating body is x = 6 cos(πt + π ) eter. What is the period, frequency and phase constant of the otion. Find the displaceent at tie t= sec. Hint: Copare given equation x = 6 cos(πt + π ) with general equation x = a cos(ω 0 t + φ) a = 6 eter, ω 0 = π and φ = π. Putting t=, we get x = 6 cos(6π + π ). 6. A spring is hung vertically and loaded with a ass of 400 g and ade to oscillated. Calculate (i) the tie-period and (ii) frequency of oscillation. When the spring is loaded with 00 g it extends by 5 c. Hint: Fro Hooke s law F = kx, where k is the force constant, x be the extension produced. Here x=5 c= 0.05, load F= 00 g=0. 9.8. k = F x = 0. 9.8 0.05 = 9.6 N/. Now T = π k = π 0.4 9.8 9.6 frequency will be f = T =.8 = 0.6 Hz. =.80 sec. Hence 7. A ass M is joined by two ass-less springs between two rigid supports, figure 4.4. If the ass is 50 g and the force constants of springs are 000 and 000 dynes/c, find(i) frequency of oscillations of the ass (ii) energy of vibration of aplitude 0.4 c (iii) velocity of ass when passing through the ean position. Solution: ass M =50 g, aplitude a= 0.4 c. let x be the displaceent fro ean position, the restoring force developed in the springs are: F = k x and F = k x, and act In the sae direction. The total restoring force F = (k + k )x, so that the force-constant of the syste is k = k + k = 000 + 000 = 5000 dynes c. Figure 4.4
Frequency of the oscillation f = π k = π 5000 50 =.6 sec. The energy of oscillation E = axiu potental energy = ka = 5000 0.4 = 400 erg = 400 0 7 Joule.Also energy E = axiu kinectic energy = v ax = 4 0 5 Joule v = 4 0 5 = 4 0. 0.05 sec 8. A rifle bullet weighing 0 g oving with a velocity of 6000 c/sec strikes and ebeds itself in a 790 g block which rests on a horizontal frictionless surface and is attached to a spring of force constant 80 Newton/. Copute the aplitude of the resulting siple haronic oscillations of block. Solution: ass bullet, = 0 g = 0.0 kg, velocity v = 6000 c. Mass of block M=790 g=0.79 kg, force constant k=8 04 dynes c conservation of linear oentu, we have: = 60 sec s = 80 Newton. According to (M + )v = v v = v = 0.0 60 =. The equation of siple haronic M+ 0.79+0.0 s otion is x = a sin ωt, where a be the aplitude and ω be the angular frequency. The velocity v = dx = aω cos ωt. At ean position, velocity is axiu i.e. ωa. dt Hence, ωa=. a = ω = T π. Since tie period T = π (M+), where k be the k force constant. Thus a = ( M+ ) = k 0.8 = 0.. 80 9. One end of a horizontal spring of force constant 6 newton/ eter is tied to a fixed wall and the other end to a solid cylinder which can roll without slipping on the horizontal ground. The cylinder is pulled a distance o. eter and released. Calculate the kinetic energy of the cylinder when passing through the equilibriu positon. Solution: Force constant k=6 newton/. The distance at which spring is stretched x= 0. eter. Hence potential energy on the spring is U = kx = 6 0. = 0.0 joule. This energy is converted in to the kinetic energy of rotation and translation of the cylinder. Hence 0.0 = Iω + v Or, 0.0 = ( r ) v + r v v = 4 0.0 = 0.04 joule. Thus kinectic energy of rotation = 4 v = 0.04 = 0.0 joule. Kinetic energy of 4 translation = 0.04 = 0.0 joule. 0. A thin rod of ass 0. kg and length 0. eter is suspended by a wire which passes through its centre and perpendicular to its length. The wire is twisted and the rod is set oscillating with a period of sec. When the flat body in the shape of equilateral triangle is suspended siilarly through its centre of ass, the period is 6 sec. Fine the rotational inertia of the triangle about the axis.
Solution: Tie period of oscillation of the rod, T = π I C.. (i) where I = Ml = 8. 0 5 kg. Siilarly tie period of the triangle (T ) = π I, C.. (ii), where I is the rotational inertia of the triangle. Dividing (i) and (ii) we get, = 0. 0. T T = I I Or, I = T I T, where T = sec and T = 6 sec. = 6 8. 0 5 4 = 7.497 0 4 kg.. Two asses o g and 90 g are connected together by a spring of length 0 c and force constant 0 6 dynes/ c. Calculate the frequency of oscillation. The ass of the spring can be neglected. Solution: The frequency of oscillation for two body oscillator is given by f = π k dynes, where k= force constant = 06 μ c = 0 N, μ = reduced ass = = 0 90 = 9 g = 0.009 kg. Thus f = 0 = 5.08 hz. + 00 π 0.009. The vibrational energy levels in HCl olecule ade fro H and Cl isotopes of atoic weights and 5 are separated by 0.6 ev. Obtain frequency and wavelength eitted in a pure vibrational transition, (ii) force constant, (iii) separation of energy levels if HCl is ade fro D and Cl 7 isotopes. Mass of hydrogen ato is.6 0 4 gra. Solution: (i) Energy separation between vibrational levels E = hν ν = 0.6.6 0 9 = 0.087 0 5 sec. The corresponding wavelength λ = c = 6.6 0 4 ν 0 8 =.44 0.087 0 5 0 6. (ii) The vibrational frequency ν = π k μ, where k be the force constant, μ be the reduced ass given by μ = H Cl = 5 = 5 5 au =.6 H + Cl 6 6 6 0 7 =.56 0 7 kg. Now k = 4π ν μ = 4.4 (0.087 0 5 ).56 0 7 = 4.65 0 N. (iii) Let μ and μ be the reduced ass of HCl and DCl respectively. Then μ = μ 5/(+5) 7/(+7) = 0.5. Let ν be the vibrational frequency corresponding to DCl. Then μ = ν μ ν ν ν = 0.5 = 0.7. Now E E = hν hν = 0.7 E = 0.7 E = 0.6 0.7 = 0.6eV.. A particle executes S.H.M. of period.4 sec and aplitude 5 c. Calculate its axiu velocity and axiu acceleration. [Ans. 0.0 /sec and 0. 0 /s ] Hint: Use v ax = ωa and a ax = ω a where a = 5 c = 0.05 and ω = π T =.4..4 4. A particle of ass 5 g executing S.H.M. has aplitude of 8 c. If it akes 6 vibrations per second, find its axiu velocity and energy at ean position. Hint: f=6 hz, ass = 5 g= 5 0 kg and aplitude a = 8c = 0.08.Now axiu velocity v = ωa = πfa = 8.0. Energy at ean position = v ax = 0.6 joule. sec
5. A particle of ass 0 g is placed in a potential field given by U = (50x +00)erg. Calculate the frequency of oscillations. Hint: F = du where U = dx (500x + 000)erg F = 000x. Now acceleration a = F = 00x. Coparing it with a = ω x, we get ω = 00 ω = 0 and πf = 0 f =.59 sec. 6. The speed v of the point P which oves in a line is given by the relation v = a + bx cx, where x is the distance of the point P fro a fixed point on the path and a,b,c are constants. Show that the otion is siple haronic if c is positive and deterine the period. Hint: We have, v = a + bx cx dv = b cx, where c is positive. Hence a = dt dv ( )x otion is S. H. M. with dt ω = c. Thus tieperiod T = π c 7. A ass of 50 g hangs fro a vertical spring of elastic constant N/. Calculate the frequency of sall vertical oscillations, assuing that daping is sall. Hint: Use f = k = 0.45 sec. π 8. A body of ass 0. kg hanging fro a spring is oscillating with a period o. second and an aplitude of eter. Find the axiu value of the force acting on the body and the value of the force constant of the spring. Hint: Here aplitude r= eter, ass =0. kg, tie period T=0. second. acceleration a ax = ω r = 985.96. Hence axiu force F = a ax = 0. 985.96 = 98.6 N. Also T = π k k = 4π T = 4.4 0. 0. = 98.6 N. 9. A body of ass 0.5 kg is projected with an energy 4 joules. If a restoring force of N/ acts on it, what axiu displaceent will the body acquire? What will be its periodic tie. Hint : F = ω x F = x ω. Hence, = 0.5 ω T = π 0.5 = π sec. If r be the aplitude, i.e. axiu displaceent, E =4 joule be the energy, then using E = ω r r = eter. 0. A block rests on a horizontal surface which is executing S.H.M. in a horizontal plane at the rate of oscillations/sec. The coefficient of static friction μ between block and plane is 0.5. How large can the aplitude be without slipping between block and surface? Hint: μ=0.5, g=9.8 /s, f= Hz, then according to question: frictional force = centripetal force. μg = ω R R = μg μg =.6 c. = ω (πf). A cylinder of ass 0.5 kg is suspended by a wire so that the axis of the cylinder reains vertical. The wire is twisted and cylinder is set oscillating. It is observed that the period is.5 sec. If this cylinder is replaced by a solid sphere of ass 0.4 kg and of sae radius as that of the cylinder. Calculate the period of oscillation of sphere. g Hint: ass of cylinder =0.5 kg, ass of sphere M=0.4 kg. Now tie period T = π I C where I be oent of inertia and C be the torsional couple per unit twist of the wire.
Here T T = I 5 0.4 r T I T = I I. But for cylinder I = r = 0.5 r and for solid sphere I = Hence.5 T =.56 T = sec.. Two bodies of ass kg and kg are joined by a assless spring of force constant k= 50 newton/eter. The bodies are pulled apart and released. Copute the frequency of the resulting oscillations. Hint: reduced ass μ = =, k=5. Then frequency of oscillation f = + 4 π k = μ π 50 =.9 Hz. /4. Find the fundaental frequency of vibration of H Cl 5 olecule. The force-constant of HCl is 5 0 5 dyne-c. Hint: Use f = π k μ,where k =5 05 dyne-c and μ= 5 +5.67 0 4 g. Hence f = 8.8 0 Hz. 4. The differential equation of a certain syste is x + τ x + ω 0 x = 0. If τω 0, find the tie at which (i) aplitude falls to /e ties the initial values (ii) energy of the syste falls to /e ties the initial values Solution: Coparing equation x + τ x + ω 0 x = 0 with standard equation of daped oscillator d x dx dt + r + ω dt 0 x = 0, we get r =. Hence the solution of this equation will τ be x = ae rt sin(ω t + φ) = ae t τ sin { (ω 0 )t + φ} (i) (ii) According to question, t a = ae τ,where a be the initial value of aplitude. e Taking log both side log e e = t log τ e e t = τ. Since energy is proportional to the square root of aplitude E a e t τ Or, E = Ca e t τ where C is a constant. If t is the tie in which the energy falls to /e ties the initial value, then e Ca = Ca e t τ. Solving t = τ. 5. If Q of a sonoeter wire is 0. On plucking, it executes 40 vibrations per second. Calculate the tie in which the aplitude decreases to of the initial value. Hint: Since Q = ω 0 τ τ = Q = 0 =.6 sec. If a be the initial aplitude, then ω 0 π 40 according to question a 0 = a e 0e rt e = e rt. Hence t = = 4τ [ τ = ] r r Or, t=4.6=5. sec. 6. For a driven haronic oscillator of natural frequency ω 0 described by π 4τ e
x + bx + kx = F 0 sin pt. Deduce an expression for the velocity aplitude x 0. Show that for the cases p > ω 0, p = ω 0 and p ω 0 this aplitude is governed by the inertia factor, resistance factor and spring factor respectively. Solution: the solution of given equation is x = sin(pt Θ) {(ω 0 p ) +4r p } Since τ = r, we get x = b sin(pt Θ). (i) where we take {(ω 0 p ) +( p τ ) } = r =, k = ω τ 0 and = F 0. Differentiating equation (i) with respect to t, we get: x = ap cos(pt Θ), where a = {(ω 0 p ) +( p τ ) } Velocity aplitude x 0 = ap = p {(ω 0 p ) +( p τ ) } CaseI : When p ω 0, velocity aplitude x 0 = p [ neglecting {p 4 +( p τ ) } ω 0 in coparision to p ]. Or, x 0 = {p +( τ ) }. Now we neglect τ in coparison to p, we get x 0 = f0 p = F 0 Hence for given driving force the velocity aplitude of the oscillator depends on inertia factor (ass). Case II: When p=ω 0, then velocity aplitude x 0 = pτ = F 0 = F 0. Hence in this case p b b velocity aplitude depends on resistance factor b f Case III: When p ω 0, Velocity aplitude x 0 = p 0 = p ω [ p ω 0 ] 0. {(ω 0 ) +( p τ ) } F0 Or, x 0 = p k = F 0p. Hence for a particular frequency p of driving force, the velocity k aplitude depends on spring factor k. p.