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PERIODIC MOTION. A repeated vibration or oscillation. Simple Harmonic Motion . Simplest form of periodic motion; it occurs when the restoring force is directly proportional to the displacement from the equilibrium For example, Hooke’s Law: F = - kx Why do we care?

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## PERIODIC MOTION

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**PERIODIC MOTION**A repeated vibration or oscillation**Simple Harmonic Motion**• Simplest form of periodic motion; it occurs when the restoring force is directly proportional to the displacement from the equilibrium • For example, Hooke’s Law: F = -kx • Why do we care? • In many cases, even if the system is not truly SHM, it can be approximated as such if the displacement is small • Tuning forks, electric current in and AC circuit, oscillations of molecules…**Describing SHM**• Amplitude (A) – maximum magnitude of displacement from equilibrium • Period (T) – time for one complete cycle • Frequency (f) – number of cycles in a unit of time, typically 1/s or hertz T = 1/f OR f = 1/T • Angular frequency (w) – 2pf – measured in radians/second**Displacement of SHM**x(t) = xmaxcos(wt + f) • x – position • xmax – amplitude • positive constant • max displacement in either direction from equilibrium • Limits of cosine are +/- 1, so limits of x are +/- x**Displacement of SHM**x(t) = xmaxcos(wt + f) • (wt + f) – phase • Time dependent • w – angular frequency • f – phase constant/phase angle • Depends on displacement and velocity of particle at t=0**Velocity of SHM**-wxmax = vmax**Acceleration of SHM**-w2xmax = amax**Force Law for SHM**F = ma F = -m(w2x) - remember that a restoring force is proportional to displacement but opposite in direction F = -(mw2)x – Hooke’s Law? (F = -kx) k = mw2**Example 1:**m = 680 kg x = 0.11 m k = 65 N/m Find: a) w, f, T b) xmax c) vmax d) amax e) What is the position function for this wave?**1980M1. A small mass m1 rests on but is not attached to a**large mass M2 that slides on its base without friction. The maximum frictional force between m1 and M2 is f. A spring of spring constant k is attached to the large mass M2 and to the wall as shown above. • Determine the maximum horizontal acceleration that M2 may have without causing m1 to slip. • Determine the maximum amplitude A for simple harmonic motion of the two masses if they are to move together, i.e., m1 must not slip on M2. • The two‑mass combination is pulled to the right the maximum amplitude A found in part (b) and released. Describe the frictional force on the small mass m1 during the first half cycle of oscillation. • The two‑mass combination is now pulled to the right a distance of A' greater than A and released. i. Determine the acceleration of m1 at the instant the masses are released. ii. Determine the acceleration of M2 at the instant the masses are released.**1990M3. A 5‑kilogram block is fastened to a vertical**spring that has a spring constant of 1,000 newtons per meter. A 3‑kilogram block rests on top of the 5‑kilogram block, as shown above. • When the blocks are at rest, how much is the spring compressed from its original length? The blocks are now pushed down and released so that they oscillate. • Determine the frequency of this oscillation. • Determine the magnitude of the maximum acceleration that the blocks can attain and still remain in contact at all times. • How far can the spring be compressed beyond the compression in part (a) without causing the blocks to exceed the acceleration value in part (c) ? • Determine the maximum speed of the blocks if the spring is compressed the distance found in part (d).**Energy and SHM**• Potential Energy for a linear oscillator depends strictly on the spring (its compression and extension) • Be careful when squaring trig functions: • cos2A = (cosA) 2 • cos A2= cos(A 2)**Energy and SHM**• Kinetic Energy for a linear oscillator depends strictly on the motion of the object**Energy and SHM**Note: at max displacement all U**Example 1 continued:**m = 680 kg x = 0.11 m k = 65 N/m Find: Energy U and K, when the block is at x=1/2 xm and x=-1/2 xm**1989M3. A 2‑kilogram block is dropped from a height of**0.45 meter above an uncompressed spring, as shown above. The spring has an elastic constant of 200 newtons per meter and negligible mass. The block strikes the end of the spring and sticks to it. a. Determine the speed of the block at the instant it hits the end of the spring. b. Determine the period of the simple harmonic motion that ensues. c. Determine the distance that the spring is compressed at the instant the speed of the block is maximum. d. Determine the maximum compression of the spring. e. Determine the amplitude of the simple harmonic motion.**1993M1. A massless spring with force constant k = 400**newtons per meter is fastened at its left end to a vertical wall, as shown in Figure 1. Initially, block C (mass mc = 4.0 kilograms) and block D (mass mD = 2.0 kilograms) rest on a horizontal surface with block C in contact with the spring (but not compressing it) and with block D in contact with block C. Block C is then moved to the left, compressing the spring a distance of 0.50 meter, and held in place while block D remains at rest as shown in Figure 11. (Use g = 10 m/s2.) • Determine the elastic energy stored in the compressed spring. Block C is then released and accelerates to the right, toward block D. The surface is rough and the coefficient of friction between each block and the surface is = 0.4. The two blocks collide instantaneously, stick together, and move to the right. Remember that the spring is not attached to block C. Determine each of the following. b. The speed vc of block C just before it collides with block D c. The speedvf blocks C and D just after they collide d. The horizontal distance the blocks move before coming to rest**Simple Pendulums**• Particle of mass “m” • Massless, unstretchable string of length “L” • Period:**Physical Pendulums**• More complicated distribution of mass • Period: where: I is the moment of inertia h is the length from the pivot point to the center of mass**Angular Oscillator**• “torsion pendulum” • Oscillates between +/- q • Restoring torque: • k is the torsion constant, dependent on the length, diameter, and material of the wire**Damped SHM**• Motion of an oscillator is reduced by an external force • F = -bv • Position function: • Energy:**Example**• M =0.25 kg; k = 85 N/m, b = 0.07 kg/s • Find T • How long does it take for xmax= 1/2 xmax • How long does it take for E= 1/2E**Forced Oscillations**• 2 frequencies associated with these systems • Natural frequency • Angular freguency of the driving force • Maximum amplitude occurs when these two frequencies are equal => resonance

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