IB Physics SL GOHS 1 13.1-13.4 Worksheet: Hooke’s Law, SHM, and Motion as a function of Time f T
Concept Questions: Fs kx 2. If a spring is cut in half, what happens to its spring constant? k a x 6. If an object–spring system is hung vertically and set into oscillation, why does m the motion eventually stop? 1 PE kx2 s 2 Problems: k Section 13.1 Hooke’s Law v (A2 x2 ) 1. A 0.40-kg object is attached to a spring with a spring constant 160 N/m so that m the object is allowed to move on a horizontal frictionless surface. The object is m released from rest when the spring is compressed 0.15 m. Find (a) the force on the T 2 object and (b) its acceleration at this instant. k 2. A load of 50 N attached to a spring hanging vertically stretches the spring 5.0 k 2f cm. The spring is now placed horizontally on a table and stretched 11 cm. What m force is required to stretch the spring by this amount? x Acos(2ft) 5. The springs 1 and 2 in Figure P13.5 have spring constants of 40.0 N/cm and 25.0 N/cm, respectively. The object A remains at rest, and both springs are stretched v A sin(2ft) equally. Determine the stretch. a A 2 cos(2ft)
Section 13.2 Elastic Potential Energy 7. A slingshot consists of a light leather cup, containing a stone, that is pulled back against two parallel rubber bands. It takes a force of 15 N to stretch either one of these bands 1.0 cm. (a) What is the potential energy stored in the two bands together when a 50-g stone is placed in the cup and pulled back 0.20 m from the equilibrium position? (b) With what speed does the stone leave the slingshot? 9. A child’s toy consists of a piece of plastic attached to a spring (Fig. P13.9). The spring is compressed against the floor a distance of 2.00 cm, and the toy is released. If the toy has a mass of 100 g and rises to a maximum height of 60.0 cm, estimate the force constant of the spring.
Section 13.3 Velocity as a Function of Position 15. A 0.40-kg object connected to a light spring with a spring constant of 19.6 N/m oscillates on a frictionless horizontal surface. If the spring is compressed 4.0 cm and released from rest, determine (a) the maximum speed of the object, (b) the speed of the object when the spring is compressed 1.5 cm, and (c) the speed of the object when the spring is stretched 1.5 cm. (d) For what value of x does the speed equal one half of the maximum speed? 17. At an outdoor market, a bunch of bananas is set into oscillatory motion with an amplitude of 20.0 cm on a spring with a spring constant of 16.0 N/m. It is observed that the maximum speed of the bunch of bananas is 40.0 cm/s. What is the weight of the bananas in newtons? 18. A 50.0-g object is attached to a horizontal spring with a spring constant of 10.0 N/m and released from rest with an amplitude of 25.0 cm. What is the velocity of the object when it is halfway to the equilibrium position if the surface is frictionless?
Section 13.4 Comparing Simple Harmonic Motion with Uniform Circular Motion 19. While riding behind a car traveling at 3.00 m/s, you notice that one of the car’s tires has a small hemispherical bump on its rim, as in Figure P13.19. (a) Explain why the bump, from your viewpoint behind the car, executes simple harmonic motion. (b) If the radius of the car’s tires is 0.30 m, what is the bump’s period of oscillation? (Hint: ω=vt/r) 21. Consider the simplified single-piston engine in Figure 13.21. If the wheel rotates at a constant angular speed ω, explain why the piston rod oscillates in simple harmonic motion.
22. The frequency of vibration of an object–spring system is 5.00 Hz when a 4.00-g mass is attached to the spring. What is the force constant of the spring? 25. A cart of mass 250 g is placed on a frictionless horizontal air track. A spring having a spring constant of 9.5 N/m is attached between the cart and the left end of the track. When in equilibrium, the cart is located 12 cm from the left end of the track. If the cart is displaced 4.5 cm from its equilibrium position, find (a) the period at which it oscillates, (b) its maximum speed, and (c) its speed when it is 14 cm from the left end of the track. 26. The motion of an object is described by the equation
Find (a) the position of the object at t = 0 and t = 0.60 s, (b) the amplitude of the motion, (c) the frequency of the motion, and (d) the period of the motion. ANSWERS:
Concept Questions: 2. Each half-spring will have twice the spring constant of the full spring, as shown by the following argument. The force exerted by a spring is proportional to the separation of the coils as the spring is extended. Imagine that we extend a spring by a given distance and measure the distance between coils. We then cut the spring in half. If one of the half- springs is now extended by the same distance, the coils will be twice as far apart as they were for the complete spring. Thus, it takes twice as much force to stretch the half-spring, from which we conclude that the half-spring has a spring constant which is twice that of the complete spring. 6. Friction. This includes both air-resistance and damping within the spring.
Problems: 13.1 (a) The force exerted on the block by the spring is
Fs = - kx = -(160 N m)( - 0.15 m) = + 24 N
or Fs = 24 N directed toward equilibrium position
(b) From Newton's second law, the acceleration is
F + 24 N m m a =s = = +60 = 60 toward equilibrium position m 0.40 kg s2 s 2
F mg 50 N 13.2 (a) The spring constant is k =s = = = 1.0 103 N m x x 5.0 10-2 m
3 2 F= Fs = kx = (1.0� 10 N m)( 0.11 m) 1.1 10 N
(b) The graph will be a straight line passing through the origin with a slope equal to k = 1.0 103 N m .
13.5 When the system is in equilibrium, the tension in the spring F= k x must equal the weight of the object. Thus,
k x (47.5 N)( 5.00 10-2 m ) k x= mg giving m = = = 0.242 kg g 9.80 m s2 13.7 (a) Assume the rubber bands obey Hooke’s law. Then, the force constant of each band is
F 15 N k =s = = 1.5 103 N m x 1.0 10-2 m
Thus, when both bands are stretched 0.20 m, the total elastic potential energy is
骣1 2 3 2 PEs =2琪 kx = ( 1.5� 10 N m)( 0.20 m) 60 J 桫2
KE+ PE = KE + PE (b) Conservation of mechanical energy gives ( s) f( s ) i , or
1 2( 60 J) mv 2 +0 = 0 + 60 J , so v = = 49 m s 2 50 10-3 kg
13.9 From conservation of mechanical energy,
1 KE+ PE + PE = KE + PE + PE 2 ( g s) ( g s ) or 0+mghf + 0 = 0 + 0 + kx i f i 2
giving
2 2mgh f 2( 0.100 kg)( 9.80 m s)( 0.600 m ) k = = = 2.94 103 N m x 2 -2 2 i (2.00 10 m )
13.15 From conservation of mechanical energy,
KE+ PEg + PE s = KE + PE g + PE s ( ) f( ) i
1 1 1 k we have mv2+0 + kx 2 = 0 + 0 + kA 2 , or v=( A2 - x 2 ) 2 2 2 m
(a) The speed is a maximum at the equilibrium position, x = 0.
k (19.6 N m ) 2 v= A 2 =(0.040 m) = 0.28 m s max m (0.40 kg ) (b) When x = -0.015 m ,
(19.6 N m ) 2 2 v =轾(0.040 m) -( - 0.015 m) = 0.26 m s (0.40 kg ) 臌
(c) When x = +0.015 m ,
(19.6 N m ) 2 2 v =轾(0.040 m) -( + 0.015 m) = 0.26 m s (0.40 kg ) 臌
1 k2 21 k 2 (d) If v= vmax , then (A- x) = A 2 m2 m
A 2 3 3 This gives A2- x 2 = , or x= A =(4.0 cm) = 3.5 cm 4 2 2
13.17 The maximum speed occurs at the equilibrium position and is
2 k kA 2 (16.0 N m)( 0.200 m ) m = = = 4.00 kg vmax = A . Thus, 2 2 , and m vmax (0.400 m s)
2 Fg = mg =(4.00 kg)( 9.80 m s) = 39.2 N
k 骣 10.0 N m 2 2 2 2 轾 13.18 v=( A - x ) =-3 (0.250 m) -( 0.125 m) = 3.06 m s m 桫琪50.0 10 kg 臌
13.19 (a) The motion is simple harmonic because the tire is rotating with constant velocity and you are looking at the uniform circular motion of the “bump” projected on a plane perpendicular to the tire.
(b) Note that the tangential speed of a point on the rim of a rolling tire is the same as the
translational speed of the axle. Thus, vt= v car = 3.00 m s and the angular velocity of the tire is
v 3.00 m s w =t = = 10.0 rad s r 0.300 m
Therefore, the period of the motion is
2p 2 p T = = = 0.628 s w 10.0 rad s 13.21 The angle of the crank pin is q= wt . Its x-coordinate is x= Acosq = A cos w t where A is the distance from the center of the wheel to the crank pin. This is of the correct form to describe simple harmonic motion. Hence, one must conclude that the motion is indeed simple harmonic.
13.22 The period of oscillations of a mass-spring system is given by T= 2p m k and the frequency is
1 1 k f = = T2p m
2 Thus, k=4p2 f 2 m = 4 p 2( 5.00 s -1) ( 4.00� 10- 3 kg) 3.95 N m
13.25 (a) The period of oscillation is T= 2p m k where k is the spring constant and m is the mass of the object attached to the end of the spring. Hence,
0.250 kg T =2p = 1.0 s 9.5 N m
(b) If the cart is released from rest when it is 4.5 cm from the equilibrium position, the amplitude of oscillation will be A =4.5 cm = 4.5 10-2 m . The maximum speed is then given by
k -2 9.5 N m vmax = Aw = A = (4.5� 10 m) 0.28 m s m 0.250 kg
(c) When the cart is 14 cm from the left end of the track, it has a displacement of x =14 cm - 12 cm = 2.0 cm from the equilibrium position. The speed of the cart at this distance from equilibrium is
k 9.5 N m 2 2 v=( A2 - x 2 ) =轾(0.045 m) -( 0.020 m) = 0.25 m s m 0.250 kg 臌
13.26 (a) At t = 0 , x =(0.30 m) cos( 0) = 0.30 m , and at t = 0.60 s ,
轾骣p x =(0.30 m) cos犏琪 rad s( 0.60 s) =( 0.30 m) cos( 0.20p rad) = 0.24 m 臌桫3 (b) A= xmax =(0.30 m)( 1) = 0.30 m
骣p (c) x= (0.30 m) cos 琪 t is of the form x= Acos (w t) with an angular frequency of 桫3 p w p 3 1 w = rad s . Thus, f = = = Hz 3 2p 2 p 6
1 (d) The period is T = = 6.0 s f