Homework Chapter 15 Solutions - Squarespace

1 Homework Chapter 15 Solutions !! ! ! ! ! ! !!page 1 Problem !A particle moving along the x axis in simple harmonic motion starts from its equilibrium position, the origin, at t = 0 and moves to the right. The amplitude of its motion is 2 cm, and the frequency is Hz.!(a)!Find an expression for the position of the particle as a function of time .!(b)!Determine the maximum speed of the particle.!(c)!Determine the earliest time (t > 0) at which the particle has this speed.

2 !(d)!Find the maximum positive acceleration of the particle.!(e)!Find the earliest time (t > 0) at which the particle has this acceleration.!(f)!Find the total distance traveled by the particle between t = 0 and t = 1 s.!Solution!(a)!The general expression for the position and velocity are! ! !At t = 0, x = 0 and v > 0.! ! !The first statement have the Solutions ! !The second statement produces a subset of the above solution.! !The angular frequency is! !The expression for the position and velocity are! ! !Use the /2 phase so that the times of the maxima are positive.

3 !(b)!The maximum speed is! !(c)!The times when with this speed is when! !(d)!The acceleration is! x(t)=Acos( t+ )v(t)= Asin( t+ )x(0)=Acos( )=0v(0)= Asin( )>0 = 2, 3 2, 5 2,.. =.., 5 2, 2,3 2,7 2,.. =2 f=2 ( 1/s)=3 1/sx(t)=(2 cm)cos(3 1/s t /2)v(t)= (3 1/s)(2 cm)sin(3 1/s t /2)vmax=(3 s 1)(2 cm)=6 cm/ssin(3 t /2)=1 3 t /2= /2 3 t= t=1/3 sa(t)= (3 1/s)2(2 cm)cos(3 1/s t /2)page 2 The maximum acceleration is! !(e)!The time of this maximum positive acceleration is! !(f)!After 1 second, the number of oscillations is!

4 !Each cycles moves through 4 amplitudes, so the total distance traveled is! !!amax=(3 1/s)2(2 cm)=18 2 cm/s2cos(3 t /2)= 1 3 t /2= 3 t=3 /2 t=1/2 sft=( cycle/s)(1 s)= cycles( cycles)(4A)=6A=6(2 cm)=12 cmpage 3 Problem !A 326 gram object is attached to a spring and executes simple harmonic motion with a period of s. The total energy of the system is J.!(a)!Find the maximum speed of the object.!(b)!Find the force constant of the spring.!(c)!Find the amplitude of the motion.!Solution!The period gives the angular frequency.

5 ! !The total energy is this which is also the maximum kinetic energy.! !(a)!The maximum speed is! !(b)!The force constant comes from! !(c)!The amplitude relates to the maximum speed.! !!T= s f=4 Hz =8 1/sE= J12mvmax2= J vmax=2( J) kg= m/s 2=km k=m 2=( kg)(8 1/s)2=206 N/mvmax= A= m/s A= m/s8 1/s= mpage 4 Problem !Show that the time rate of change of mechanical energy for a damped, un-driven oscillator is given by! !and hence is always negative. To do so, differentiate the expression for the mechanical energy of an oscillator.

6 !Solution!The energy is! !Its derivative is! !The differential equation that describes this system is! !The first term in the derivative above is! ! !!dEdt= bv2E=12mv2+12kx2dEdt=12m(2vdvdt)+12k(2xd xdt)=mva+kxvma= kx bvmva= kxv bv2dEdt= kxv bv2+kxv= bv2page 5 Problem !A pendulum of length L and mass M has a spring of force constant k connected to it at a distance of h below its point of suspension. Find the frequency of vibration of the system for small values of the amplitude.!Solution!I will approach this through torque.

7 ! !The torque from gravity is! !The torque from the spring is! !The displacement from equilibrium is related to the angle this way.! ! !Both of these torques act in unison so they have the same sign. The net torque is! !Expanding the angular functions for small angles,! !The net torque is! !This means the angular frequency is! !!LMg hkx + /2 gravity=L Mg sin( ) spring=h kx sin( + /2)=h kx cos( )x=hsin spring=h khsin( ) cos( ) =MgLsin( )+h2ksin( )cos( ) =(MgL+h2k) ML2d2 dt2=(MgL+h2k) d2 dt2=(MgL+h2k)ML2 =(MgL+h2k)ML2 f=12 (MgL+h2k)ML2page 6 LhMk Problem !

8 An object of mass m1 = 9 kg is in equilibrium when connected to a light spring of constant k = 100 N/m that is fastened to a wall. A second object m2 = 7 kg is slowly pushed up against m1 compressing the spring by the amount A = m. The system is then released and both objects start moving to the right on the frictionless surface.!(a)!When m1 reaches the equilibrium position, m2 loses contact with m1 and moves to the right with speed v. Determine the value of v.!(b)!How far apart are the objects when the spring is fully stretched for the first time ?

9 !Solution!Between the initial position and the equilibrium position, the spring acts on both masses together. The motion is determined by this.! !The amplitude is m. The initial position is m.! !The angular frequency is! !The motion is! ! !At the equilibrium position, the time is! ! !At this time , the speed of both masses is! !(b)!After the separation, the motion of the still-connected mass m1 changes. The angular frequency is! !Reseting t = 0 s to the equilibrium position.! !The velocity is! !At t = 0 s, the speed is m/s.

10 ! !x(t)=Acos( t+ ) ( ) =.. 3 , , ,3 ,.. =km=100 N/m16 kg= 1/sx(t)=( m)cos( 1/s t )v(t)= ( 1/s)( m)sin( 1/s t )0=( m)cos( 1/s t ) 1/s t = /2 /2t= /5= sv(t)= ( 1/s)( m)sin( 1/s s )= m/s =km=100 N/m9 kg= 1/s0=Acos( ) =.., 3 2, 2, 2,3 2,..v(t)= ( 1/s)Asin( 1/s t /2) ( 1/s)Asin( /2) A= mpage 7 The motion of m1 is! ! !It takes this much time for m1 to reach the fully stretched position of m.! !After this much time , the mass m2 has traveled a distance of ! !The difference between the two masses position is m = cm.