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Transcript
Forces in a Spring If a spring is strained so that its length deviates from its equilibrium value by x (positive x means extension, negative x means compression), Hook’s law says that a recoil force, F exists in the spring and is given by: Simple Harmonic Motion Unstrained or equilibrium length of the spring x is positive here. x is negative here. F = −k x Negative because the recoil force always tries to restore the spring to its equilibrium length or to reduce x. 1 where k is the spring constant, a measure of the stiffness of the spring. It has units of N/m. Example 1 Spring Constant of a Spring. A spring hangs vertically down from a support. When you hang a 100-gram mass from the bottom end of the spring and stop any motion of the system the spring is stretched by 10 cm. Determine the spring constant. Hint: sketch a freev v body diagram and use F = −k x If we hang another 100-gram mass on the spring, the spring stretches further. Is the additional stretch more than, less than, or equal to 10 cm? Energy Stored in a Spring The elastic potential energy of a strained spring is : U= 1 2 kx 2 Note that the force in a spring is a kind of conserved force because the work done by the spring depends only on x and is zero when the spring returns to its starting position. Therefore, we call the energy in a spring a potential energy as we do for the energy associated with the gravitational force. 3 Question: A block connected to a horizontal spring sits on a frictionless table. The system is released from rest, with the spring initially stretched. What happens to the energy stored in the spring? What is the maximum speed of the block? Solution: As the block oscillates, the energy goes back and forth between elastic potential energy and kinetic energy. Simulation ⇒ ⎛ k ⎞ v max = xmax ⎜ ⎟ ⎝ m⎠ 1 2 1 2 kxmax = mv max 2 2 Note that initially the block is at rest. Hence Ui = Umax and correspondingly xi = xmax. 4 Example 3 Maximum Compression of a Spring. Example 2 Energy in an Oscillating Spring. Umax = K max 2 Question: A block with a mass of 1.0 kg is released from rest on a frictionless incline. At the bottom of the incline, which is 1.8 m vertically below where the block started, the block slides across a horizontal frictionless surface before encountering a spring that has a spring constant of 100 N/m. (a) What is the maximum compression of the spring? (b) What is the maximum acceleration of the block after it makes contact with the spring? Solution: (a) Ui = Uf ⇒ mgh = xmax = 5 2mgh = k 1 2 kxmax 2 2 × 1.0 kg × 10 N/kg × 1.8 m = 100 N/m = 0.36 m2 = 0.60 m 6 1 Example 3 Maximum Compression of a Spring (cont’d). Simple Harmonic Motion (SHM) (b) The acceleration is the maximum when the force is maximized, which is when the spring is the most compressed. Here, we only need to be concerned with the magnitude of the variables. The oscillatory motion we saw in the last examples is call simple harmonic motion, in which no energy is lost. The energy only gets converted between elastic potential energy and kinetic energy during the motion. Fmax = kx = 100 N/m × 0.60 m = 60 N amax = Fmax 60 N = = 60 m/s2 m 1.0 kg An illuminative way to see how the different kinetic variables, x, v and a evolve with t in a simple harmonic motion is to use a reference circle (to be elicited in the next four pages). 7 8 Displacement, x Displacement in a SHM 1 cycle +A The displacement, x, of the shadow of a ball undergoing uniform circular motion conforms to that of a SHM. t x = A cosθ = A cos ωt Film (top x=0m view) where ω = angular frequency, A = amplitude of the SHM. −A Period = T Several characteristic variables of a SHM: Position at t = 0 Amplitude, A: The maximum displacement Reference circle Period, T: The time required to complete one cycle Frequency, f: The number of cycles per second (measured in Hz) f = 9 Light 1 T ω = 2π f = 2π T 10 Acceleration in a SHM Velocity in a SHM v x = −vT sin θ = − { Aω sin ωt vmax a x = − ac cos θ = − { Aω 2 cos ωt amax a = −ω2x amax = Aω2 v = −Aω sinωt vmax = Aω 11 12 2 x = A cos(ω t ) Motion Graphs What determines the angular frequency? v From last page, a = −ω 2 xv . If we graph position, velocity, and acceleration of the object on the spring, as a function of time, we get the following. The period of these oscillations happens to be 4.0 seconds. v = − Aω sin(ω t ) In turn, by Newton’s second law, we have: v v ∑ F = ma v v −kx = ma v k v a=− x m a = − Aω 2 cos(ω t ) ω2 = k m Since vmax = k m 14 We have an object (mass m) attached to a massless spring. The object is on a horizontal frictionless surface. We move the object so the spring is stretched, and then we release it. The object oscillates back and forth. (3) A2ω2 ω= Understanding Oscillations from Energy Graphs Etot = U(t) + K(t) = Umax = Kmax 2 ⇒ Hence ω is determined exclusively by k and m only. What determines the energy of a SHM? Kmax = ½ mvmax2 (2) By (1) and (2), we have: 13 Umax = ½ kA2 (1) (4) = A2(k/m), (3) and (4) are equivalent. To understand the motion, let’s take a look at graphs of kinetic energy and elastic potential energy, first as a function of time and then as a function of position. Which color goes with kinetic energy, and which with elastic potential energy? From the above, we can conclude that the energy of a SHM is determined by k and A (by eqn. (3)) or equivalently m and vmax. Simulation 15 Which graph is which? 2. 3. 1 2 kx 2 K = ½ mv2 16 Example 4 Elapsed time in a SHM. Energy graphs 1. U= The red one is the kinetic energy; the blue one is the potential energy. The blue one is the kinetic energy; the red one is the potential energy. The graphs are interchangeable so you can't tell which is which. An object attached to a spring is pulled a distance A from the equilibrium position and released from rest. It then experiences simple harmonic motion with a period T. The time taken to travel between the equilibrium position and a point A from equilibrium is T/4. How much time is taken to travel between points A/2 from equilibrium and A from equilibrium? Assume the points are on the same side of the equilibrium position, and that mechanical energy is conserved. 1. T/8 2. More than T/8 3. Less than T/8 4. It depends whether the object is moving toward or away from the equilibrium position 17 18 3 Example 4 Elapsed time in a SHM (cont’d) Suppose that at t = 0 the object is at x = A from equilibrium. So the equation, x = A cos(ω t ) applies. x = A cos(ω t ) Let’s solve for the time t when the object is at x = A/2 from equilibrium. A 1 = A cos(ω t ) ⇒ = cos(ω t ) 2 2 2π By using ω = , the above equation becomes: T 1 2π t = cos( ) T 2 By taking the inverse cosine of both sides, we get: t π 3 19 Another Example of SHM -- Pendulum = 2π t T ⇒ t= T 6 (remember to express the angles in radian) This is more than T/8, because the object travels at a small average speed when it is far from equilibrium. 20 Take torques around the support point. v v ∑ τ = Iα Free-body diagram of a pendulum when it is displaced to the left. −Lmg sinθ = mL2α g L α = − sinθ If angle θ is small, sinθ ≈ θ Free-body diagram of a pendulum when it passes through equilibrium. ⇔ α = −ω 2θ g α ≈ − θ , which has the SHM form a = −ω2x (see eqn. (1) L on p. 14) Simulation 21 So, the angular frequency is ω = g L 22 Damped Harmonic Motion In simple harmonic motion, an object oscillated with a constant amplitude. In reality, friction or some other energy dissipating mechanism is always present and the amplitude decreases as time passes. 1) simple harmonic motion This is referred to as damped harmonic motion. 2&3) underdamped 4) critically damped 5) overdamped 23 24 4 Driven Harmonic Motion Resonance When a force is applied to an oscillating system at all times, the result is driven harmonic motion. Resonance is the condition in which a time-dependent force can transmit large amounts of energy to an oscillating object, leading to a large amplitude motion. Here, the driving force has the same frequency as the spring system (i.e., √(k/m)) and always points in the direction of the object’s velocity. Resonance occurs when the frequency of the force matches a natural frequency at which the object will oscillate. 25 26 5