University Physics AI No. 5 Simple Harmonic Oscillation Class Number Name I. Choose the Correct Answer 1. A particle on a spring executes simple harmonic motion. If the mass of the particle and the amplitude are both doubled then the period of oscillation will change by a factor of ( D ) (A) 4. (B) 8 . (C) 2. (D) 2 Solution: Since the period of oscillation is k m T π ω π 2 2 == , according to the problem T k m k m k m T 222 2 22 =?== ′ =′ πππ 2. A particle on a spring executes simple harmonic motion; when it passes through the equilibrium position it has a speed v. The particle is stopped, and then the oscillations are restarted so that it now passes through the equilibrium position with a speed of 2v. After this change the frequency of oscillation will change by a factor of ( E ) (A) 4. (B) 8 . (C) 2. (D) 2 (E) 1 Solution: Since the frequency of oscillation is m k ππ ω ν 2 1 2 == , the frequency of oscillation will not change. 3. A particle on a spring executes simple harmonic motion; when it passes through the equilibrium position it has a speed v. The particle is stopped, and then the oscillations are restarted so that it now passes through the equilibrium position with a speed of 2v. After this change the maximum displacement will change by a factor of ( C ) (A) 4. (B) 8 . (C) 2. (D) 2 (E) 1 Solution: The energy of simple harmonic motion is pk EEkAE +== 2 2 1 When the oscillation passes through the equilibrium position, its energy is 2 2 1 mvEE k == . So we have k m vA = . According to the problem, A k m vA 22 ==′ 4. A particle on a spring executes simple harmonic motion. When the particle is found at x=x max /2 the speed of the particle is ( B ) (A) max vv x = (B) 2/3 max vv x = (C) 2/2 max vv x = (D) 2/ max vv x = Solution: The energy of simple harmonic motion is pk EEkAE +== 2 2 1 When the oscillation is at the equilibrium position, its energy is 2 max 2 1 mvEE k == When the oscillation is at the maximum displacement, its energy is 2 max 2 1 kxEE p == So we have 2 max 2 max 2 2 1 2 1 2 1 kxmvkAE === According to the problem, max 2 max 22 max 2 max 2 2 3 2 1 4 1 2 1 2 1 ) 2 ( 2 1 2 1 vvmvmvmv x kmvE =??+=?+= 5. A particle on a spring executes simple harmonic motion. If the total energy of the particle is doubled then the magnitude of the maximum acceleration of the particle will increase by a factor of ( D ) (A) 4. (B) 8 . (C) 2. (D) 2 (E) 1 (it remains unchanged). Solution: The acceleration of the particle is )cos( 2 φωω +?= tAa , Its maximum acceleration is 2 max ωAa = . The energy of simple harmonic motion is 2 2 1 kAE = According to the problem AAAkkAEE 2 2 1 2 1 22 22 =′?′=?==′ Then the maximum acceleration is 22 max 2 ωω AAa =′=′ II. Filling the Blanks 1. Two springs with spring constants k 1 and k 2 are connected as shown in Figure 1 with a mass m attached on a frictionless surface. We pull on m with a force we F r and hold m at rest. The mass m moves a distance x and point A moves a distance x′ from their equilibrium positions. By examining the forces on point A, you can get the relation of x and x′ is x kk k x 21 2 + =′ . If the system is modeled with a single spring stretching m a distance x, by examining the forces on m, the effective spring constant of the single spring is 21 21 kk kk + . Solution: According to the problem, we have x kk k xxxkxk xxkF xkF FFF k k wekk 21 2 21 2 1 )( )( 2 1 21 + =′?′?=′? ? ? ? ? ? ? ? ′?= ′= == 21 21 21 2 11 ' 21 kk kk kkxx kk k kkxxkFFF wekk + =?= + ??=?== 2. A 500 g mass is undergoing simple harmonic oscillation that is described by the following equation for its position x(t) from equilibrium: ]rad 0.6 )rad/sπ0.6cos[()m50.0()( π += ttx The amplitude A of the oscillation is 0.5m . The angular frequency ω of the oscillation is 6.0π rad/s . The frequency ν of the oscillation is 3 Hz . The period T of the oscillation is 1/3 s . Fig.1 Frictionless Surface m k 1 k 2 A i ? The spring constant k is 177.47 N/m . The position of oscillator when t = 0 s is 0.43 m . The time (t > 0 s ) the oscillator first at maximum distance from equilibrium is 0.44 s . The maximum speed of the oscillator is 9.42 m/s . The magnitude of the maximum acceleration of the oscillation is 177.47 m/s 2 . Solution: From the equation ] 0.6 rad )rad/sπ0.6cos[()m50.0()( π += ttx , we get (a) m5.0=A (b) rad/s6πω = (c) Hz3 2 6 2 === π π π ω ν (d) s 3 1 6 22 === π π ω π T (e) N/m47.177185.0)6( 222 ==×==?= ππωω mk m k (f) When 0=t m433.0 4 3 6 cos5.0 === π x g) π π π π π π π ktttx =+?±=+?=+= 6 61) 6 6cos(5.0) 6 6cos(5.0 stk kkt 44.0 36 5 1If ,2,1,0) 6 ( 6 1 === ±±=?=? π π π π K (h) ) 6 6cos(5.0 π π += tx ) 6 6sin(3 d d π ππ +?== t t x v m/s42.93 max == πv (i) ) 6 6cos(18 d d 2 π ππ +?== t v a 22 max m/s47.17718 == πa 3. An oscillator consists of a block attached to a spring (k=456N/m). At some time t, the position (measured from the equilibrium location), velocity, and acceleration of the block are x = 0.112m, v x =-13.6m/s, a x =-123m/s 2 . The frequency of the oscillation is 5.28Hz , the mass of the block is 0.42kg , the amplitude of oscillation is 0.43 m . Solution: The general equation about the oscillation is ? ? ? ? ? +?= +?= += )3()cos( )2()sin( )1()cos( 2 φωω φωω φω tAa tAv tAx Substitute x, v, a to the euqtions, solving (1) and (3), we have )mN/kg(14.33 ?=ω So the frequency of the oscillation is )Hz(28.5 14.32 14.33 2 = × == π ω f The mass of the block is kg42.0 14.33 456 22 === ω k m Solving (2) and (3), we have the amplitude of oscillation is m43.0112.0 14.33 6.13 2 2 2 2 2 2 =+=+= x v A ω 4. A 10.0 kg mass attached to a spring is dragged at constant speed up rough surface (μ s = 0.30, μ k = 0.25) inclined at 10° to the horizontal as in Figure 2. The spring is stretched 8.0 cm. The spring constant of the spring is 514.32 N/m . If the spring drags the mass at constant speed down the slope, the stretch of the spring is 0.014m . Solution: The second law force diagrams of the mass are shown in figure. (a) Up the slop Apply Newton’s second law of motion, we have ? ? ? ? ? ? ? ?= = =? =?? xkF Nf mgN mgfF k μ 010cos 010sin o o F r Fig.2 10° f v F r gm v N v )N/m(32.514)10sin10cos( =+ ? =? oo k x mg k μ (b) Down the slop Apply Newton’s second law of motion, we have )m(014.0)10sin10cos( 010cos 010sin =?=?? ? ? ? ? ? ? ? ′?= = =? =+? oo o o k k k mg x xkF Nf mgN mgfF μ μ III. Give the Solutions of the Following Problems 1. A 1.50 kg mass on a horizontal frictionless surface is attached to a horizontal spring with spring constant k = 200 N/m. The mass is in equilibrium at x = 0 m. The mass is released when t = 0 s at coordinate x = 0.100 m with a velocity (2.00 m/s) i ? . (a) Find the constants A, ω and φ in )cos()( φω += tAtx . (b) Find the period T of the oscillation. (c) Determine the maximum speed of the oscillation and the magnitude of the maximum acceleration. (d) Plot x(t) during the time interval t = 0 s and t = 2T. Solution: (a) From the problem, mN/kg3.133 5.1 200 ?=== m k ω , so substitute A, ω, t= 0 s, ,and x = 0.100 m and v=2.00 m/s to )cos()( φω += tAtx , we have ? ? ? ?=?== == φφω φ sin3.133sin2 cos1.0 0 0 AAv Ax Solving the equations, we have ? ? ? ? ? ? ? ?=?= =+= 33 5 ) 3.133 arctan( )m(1.0) 3.133 2 (1.0 0 0 22 π πφ or x v A Thinking of the direction of the velocity of the mass at t= 0 s, we get 3 π φ ?= . N v f v F r gm v (b) The period T of the oscillation is )s(54.0 3.133 22 === π ω π T (c) The maximum speed of the oscillation is m/s33.133.1331.0 max =×== ωAv The magnitude of the maximum acceleration is 2322 max m/s1078.13.1331.0 ×=×== ωAa 2. A 5.13kg object moves on a frictionless surface under the influence of a spring with force constant 9.88N/cm. The object is displaced 53.5cm and given an initial velocity of 11.2m/s back toward the equilibrium position. Find (a) the frequency of the motion, (b) the initial potential energy of the system, (c) the initial kinetic energy, and (d) the amplitude of the motion. Solution: (a) The frequency of the motion is )Hz(21.2 2 1 2 === m k ππ ω ν (b) The initial potential energy of the system is )J(4.141535.0988 2 1 2 1 22 =××== kxPE (c) The initial kinetic energy of the system is )J(73.282.1113.5 2 1 2 1 22 =××== mvKE (D) As the total kinetic energy is 2 2 1 kAKEPEE =+= , Thus the amplitude of the motion is )m(59.0 988 )73.284.141(2)(2 = +× = + = k KEPE A 3. Electrons in an oscilloscope are deflected by two mutually perpendicular electric forces in such a way that at any time t the displacement is given by tAx ωcos= and )cos( y tAy φω += . Describe the path of the electrons and determine its equation when (a) o 0= y φ , (b) o 30= y φ , and (c) o 90= y φ . Solution: (a) When o 0= y φ , tAx ωcos= , tAy ωcos= , Thus y=x, it is a straight line. (b) When o 30= y φ , ? ? ? ? ? += = ) 6 cos( cos π ω ω tAy tAx , Solving it, we get 222 4 1 3 Axyyx =?+ , it is a ellipse. (c) When o 90= y φ , ? ? ? ? ? += = ) 2 cos( cos π ω ω tAy tAx Solving it, we get 1 2 2 2 2 =+ A y A x , it is a circular. 4. Two masses are undergoing simple harmonic oscillation according to the equations x 1 (t)=(0.150m) cos [(0.250 rad/s) t + π/6 rad] x 2 (t)=(0.200m) cos [(0.350 rad/s) t + π/4 rad] (a) Find the period of each oscillation. (b) On the same graph, plot x 1 (t) and x 2 (t) versus t during t = 0 s to 60 s and determine the first six instants at which they have the same position. Is the interval between these instants the same? (c) By changing one of the amplitudes, see if the instants when the oscillators have the same position depend on the amplitudes. Solution: (a) s12.258 25.0 22 1 1 ==== π π ω π T s94.17 7 40 35.0 22 2 2 ==== ππ ω π T (b) T=0.36 s x 1 = x 2 =0.12 m 9.01 s -0.14 m 29.22 s 0.004 m 39.43 s 0.09 m 49.40 s 0.14 m 57.78 s -0.11 m The interval is not same. (c) The instants when the oscillation has the same position depend on the amplitudes. 5. A spring with spring constant k is attached to a mass m that is confined to move along a frictionless rail oriented perpendicular to the axis of the spring as indicated in Figure 3. The spring is initially unstretched and of length l 0 when the mass is at the position x = 0 m in the indicated coordinate system. Show that when the mass is released from the point x along the rail, the oscillations occur but their Fig.3 l 0 k x m θ O oscillations are not simple harmonic oscillations. Solution: The mass is pulled out a distance x along the rail, the new total length of the spring is 2 2 0 xll += So the x-component of the force that the spring exerted on the mass is θθ cos])[(cos 0 2 1 2 2 0 lxlkFF x ?+?== From the diagram , 2 0 2 cos lx x + =θ , so the x-component of the force is ) 1)( 1 1()( 2 0 2 0 2 0 + ??= + ??= l x kx lx xl xkF x ) 1)( 1)( 1( 2 0 2 0 + + ??= l x l x kx For 1)( 2 0 0 <<?<< l x lx , the x-component of the force is ]1)(1[ 2 0 +??≈ l x kxF x Since L??=+ 2 0 2 0 )( 2 1 11)( l x l x Then 3 2 0 2 0 2 2 ) 2 1 11( x l k l x kxF x ?=?+??≈ Thus when the mass is released from the point x along the rail, the oscillations occur but their oscillations are not simple harmonic oscillations.