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Transcript
The Spring ❑Consider a spring, which we apply a force FA to either stretch it or compress it FA x unstretched -FA -x x=0 FA = kx k is the spring constant, units of N/m, different for different materials, number of coils From Newton’s 3rd Law, the spring exerts a force that is equal in magnitude, but opposite in direction Hooke’s Law for the restoring force of an ideal spring. (It is s a conservative force.) F = −kx Chapter 4, problem P29 Five identical springs, each with stiffness 390 N/ m, are attached in parallel (that is side-by-side) to hold up a heavy weight. If these springs were replaced by an equivalent single spring, what should be the stiffness of this single spring? Oscillatory Motion ❑ We continue our studies of mechanics, but combine the concepts of translational and rotational motion. ❑ In particular, we will re-examine the restoring force of the spring (later its potential energy). ❑ We will consider the motion of a mass, attached to the spring, about its equilibrium position. ❑ This type of motion is applicable to many other kinds of situations: pendulum, atoms, planets, ... Simple Harmonic Motion ❑ If we add a mass m to the end of the (massless) spring, stretch it to a displacement x0, and release it. The spring-mass system will oscillate from x0 to –x0 and back. Without friction and air m resistance, the oscillation would k continue indefinitely -x0 0 x0 ❑ This is Simple Harmonic Motion (SHM) ❑ SHM has a maximum magnitude of |x0|=A, called the Amplitude ❑ One way to understand SHM is to consider the circular motion of a particle and rotational kinematics (The Reference Circle) ❑ The particle travels on a circle of radius r=A with the line from the center to the particle making an angle θ with respect to the x-axis at some instant in time • Now, project this 2D motion onto a 1D axis x = A cosθ but θ = ω t x=-A x=0 x=A x Therefore, x = A cos(ω t ) ❑ x is the displacement for SHM, which includes the motion of a spring A θ A ❑ SHM is also called sinusoidal motion ❑ xmax=A=x0=amplitude of the motion (maximum) • ω is the angular frequency (speed) in rad/s. It remains constant during the motion. ❑ ω and the period T are related • Define the frequency • 2π ω= T # cycles 1 Units of 1/s = Hertz (Hz) f = = sec T 2π Relates frequency and ω= = 2π f angular frequency T • As an example, the alternating current (AC) of electricity in the US has a frequency of 60 Hz • Now, lets consider the velocity for SHM, again using the Reference circle v = -v t sin θ v t = rω = Aω v = -Aω sin(ω t ) ❑ Amplitude of the velocity is v max = A ω x vt v θ A ❑ Acceleration of SHM x a = −ar cosθ 2 2 2 vt r ω ar = = r r a ar θ 2 2 2 ar = rω = Aω ⇒ a = − Aω cos(ω t ) ❑ The amplitude of the acceleration is amax = Aω ❑ Summary of SHM 2 xmax = A, ω t = 0, π ,2π ,... x = A cos(ω t ) π 3π v = − Aω sin(ω t ) v max = Aω , ω t = , ... 2 2 2 a = − Aω cos(ω t ) a = Aω 2 , ω t = 0, π ,2π ,... max x (A=1) ωt (π rads) v (Aω=1) ωt (π rads) a (Aω2=1) Example Problem Given an amplitude of 0.500 m and a frequency of 2.00 Hz for an object undergoing simple harmonic motion, determine (a) the displacement, (b) the velocity, and (c) the acceleration at time 0.0500 s. Solution: Given: A=0.500 m, f=2.00 Hz, t=0.0500 s. (a) ω = 2πf = 2π (2.00 Hz) = 4.00π rad/s ωt = (4.00π rad/s)(0.0500 s) = 0.200π rad = 0.628 rad x = A cos(ωt ) = (0.500 m)cos(0.628 rad) x = 0.405 m v = - Aω sin(ω t ) v = −(0.500 m)(4π rad/s) sin(0.628 rad) v = −3.69 m/s 2 a = Aω cos(ω t ) 2 a = −(0.500 m)(4π rad/s) cos(0.628 rad) 2 a = −63.9 m/s Frequency of Vibration ❑ Apply Newton’s 2nd Law to the spring-mass system (neglect friction and air resistance) k m 0 Fs x A ∑F = ma x Consider x-direction only Fs = −kx = ma x Substitute x and a for SHM 2 − k[ A cos(ω t )] = m[− Aω cos(ω t )] 2 k = mω k Angular frequency of vibration for a ω= spring with spring constant k and m attached mass m. Spring is assumed x to be massless. 1 ω = 2π f ⇒ f = 2π 1 m T = = 2π =T f k k m Frequency of vibration Period of vibration ❑ This last equation can be used to determine k by measuring T and m m 2 m T = 4π ⇒ k = 4π 2 k T 2 2 ❑Note that ω (f or T) does not depend on the amplitude of the motion A • Can also arrive at these equations by considering derivatives The Simple Pendulum ❑ An application of Simple Harmonic Motion ❑ A mass m at the end of a massless rod of length L ❑ There is a restoring force which acts to restore the mass to θ=0 L θ F = −mg sin θ m • Compare to the spring Fs=-kx • The pendulum does not display SHM T mgsinθ mg ❑ But for very small θ (rad), we can make the approximation (θ<0.5 rad or about 25°) → simple pendulum approximation sin θ ≈ θ ⇒ F = −mgθ This is SHM since s = rθ = Lθ Arc length s mg Looks like spring F = −mg = − s force L L mg Like the spring ( Fs = −kx) ⇒ k = constant L ❑ Now, consider the angular frequency of the spring k mg / L g ω= = = =ω m m L 1 g Simple pendulum f = 2π L frequency Simple pendulum angular frequency • With this ω, the same equations expressing the displacement x, v, and a for the spring can be used for the simple pendulum, as long as θ is small • For θ large, the SHM equations (in terms of sin and cos) are no longer valid → more complicated functions are needed (which we will not consider) • A pendulum does not have to be a point-particle