Phy 202: General Physics II Ch 10: Simple Harmonic Motion & Elasticity

The Ideal Spring & Hooke’s Law Springs are objects that exhibit elastic behavior An ideal spring is: –Massless (the mass of the spring is negligible compared to –The applied force (F applied ) required to compress/stretch is proportional to the displacement of the spring from its unstrained length (x) or F applied = kx where k is called the spring constant (or stiffness of the spring) –To stretch/compress a spring, the spring exerts a restoring force of equal & opposite magnitude (reaction force, F) against the stretching/compressing force or F = -kx {this is referred to as Hooke’s Law!}

The (Elastic) Restoring Force (& Newton’s 3 rd Law) Action: –Applied force is proportional to displacement of the spring: F applied = kx Reaction: –Restoring force is equal/opposite to applied force: F = -F applied = -kx

Work Performed by an Ideal Spring When an ideal spring is stretched a displacement x by an applied force, the average force applied to the spring is F avg = ½ kx The work performed to stretch the spring is W = (F avg cos  )x = ½ kx 2 The strained spring therefore gains an elastic potential energy PE elastic = ½ kx 2

Displacing an Ideal Spring results in Simple Harmonic Motion Releasing a strained a spring results in oscillating motion due to the spring restoring force Motion oscillates between (+x and –x) This type of motion is called Simple Harmonic Motion

Newton’s 2 nd Law & Ideal Springs Applying Newton’s 2 nd Law to a stretched ideal spring:  F = ma = -kx The acceleration of the spring is a = - (k/m). x The acceleration of the spring at any point in the motion is proportional to the displacement of the spring For motions of this type, the angular frequency (  ) of the motion is  =(k/m) ½ The period of the motion (T) is T = 1/f = 2  /  = 2   (m/k) ½ General form: a =  2 x ( when a ~ x)

Simple Harmonic Motion When the restoring force of a spring obeys Hooke’s Law (F=-kx), the resulting motion is called Simple Harmonic Motion Consider a mass attached to a stretched spring that is released at t o =0: –The displacement (x) of the mass due to the spring’s restoring force will be x = A cos  t where A is the amplitude of the strained spring –The velocity (v) of the mass will be v = -A  sin  t = -v max sin  t where v max = A  –The acceleration (a) of the mass will be a = -A  2 cos  t = -a max cos  t where a max = A  2

Graphical Perspective of SHO Displacement (x) x = Acos(  t) Velocity (v) v = -v max sin(  t) Acceleration (a) a = -a max cos(  t)

Conservation of Energy & Simple Harmonic Motion When work is performed on a spring due to stretch/compression by an applied force the spring gains potential energy equal to PE elastic = ½ kx 2 As the spring is released and the restoring force w/in the spring drives the motion of the spring (assuming no friction) –PE elastic is converted to KE as the spring force does work –When the spring’s length equals its unstretched length, all of the PE elastic is converted to KE Applying conservation of Energy to the spring: (PE elastic ) stretched = KE unstretched or ½ kx 2 = ½ mv 2 Therefore, the speed of the spring at its unstretched length is related to the length of the original displacement of the spring: v = (k/m) ½

The Pendulum Consider the motion of a mass (m) attached to a string (length, l): –The gravitational force (mg) exerts a torque on the mass (at all but the bottom point of the swing)  = I  = mgl sin  since I = ml 2 and sin  ~   = (mgl/I) sin  = (mgl/ml 2 ) sin   = (g/l). sin  or  = (g/l).  –The period of the motion (T) is therefore T = 2  (l/g) ½ Note:  ~  similar to a ~x (for a spring) therefore this motion looks like the form:  =  2  l m  mg

Elastic Deformation Types of deformation: –Stretching/compression F stretch/compress = Y(  L /L o )A –Shear deformation F shear = S (  X/L o )A –Volume deformation  P = -B (  V/V o )

Stress, Strain & Hooke’s Law We can consider a strained mass as though it were a collection of small masses attached by a system of springs Since deformation is related to the applied force: F stretch/compress = (YA/L o )  L The effective spring constant (k effective ) for the mass is k effective = YA/L o Note that k effective is inversely proportional to L o Question: Consider a long spring (spring constant=k). How do the spring constants of the smaller pieces (k’) compare to the original k?