The energy of simple harmonic motion is E==kA2 According to the problem E"=2E=2·kf2=kr2→A=√2A Then the maximum acceleration is a=402=v2A02 I. Filling the blanks 1. Two springs with spring constants k, and k2 are Frictionless connected as shown in Figure I with a mass m attached on a frictionless surface. We pull on m with a (00000 ook Surface force F 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 xis xk.+kx. 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 k, k2 k1+k2 According to the problem, we have F=F =F f =kx k, x'=k,(x-x) (x-x F.=F,=F→kx=kx→k1 2. A 500 g mass is undergoing simple harmonic oscillation that is described by the following equation for its position x(o) from equilibrium x((=(0.50m)cos(6.0 rad/s)t+2-rad The amplitude A of the oscillation is_0.5m_. The angular frequency o of the oscillation is 6.0 rad/s. The frequency v of the oscillation is_3 Hz The period Tof the oscillation is_ 1/3s.The energy of simple harmonic motion is 2 2 1 E = kA According to the problem E E kA kA A 2A 2 1 2 1 2 2 2 2 ′ = = ⋅ = ′ ⇒ ′ = Then the maximum acceleration is 2 2 max a′ = A′ω = 2Aω II. Filling the Blanks 1. Two springs with spring constants k1 and k2 are connected as shown in Figure 1 with a mass m attached on a frictionless surface. We pull on m with a force Fwe 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 k k k x 1 2 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 1 2 1 2 k k k k + . Solution: According to the problem, we have x k k k k x k x x x F k x x F k x F F F k k k k we 1 2 2 1 2 2 1 ( ) ( ) 2 1 1 2 + ⇒ ′ = − ′ ⇒ ′ = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = − ′ = ′ = = 1 2 1 2 1 2 2 1 1 ' 1 2 k k k k x kx k k k k F F F k x kx k k k we + = ⇒ = + = = ⇒ = ⇒ ⋅ 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] 6.0 ( ) (0.50m)cos[(6.0π rad/s) π x t = t + 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 k1 A k2 i ˆ