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Chapter 10

Younes Sina's Lecture at Pellissippi State Community College

Younes Sina's Lecture at Pellissippi State Community College

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Chapter 10

1. 1. Chapter 10 Simple Harmonic Motion (SHM): Younes Sina
2. 2. A mass attached to a linear spring and set into up-and-down motion performs a motion that is called " simple harmonic motion (SHM). Linear Springs: A linear spring is one for which the change in length ( Δx ) is proportional to the change in the applied force ( ΔF ). ΔF = k Δx spring constant The Metric unit for k is N/m
3. 3. Example : A linear spring has an unstretched length of 18 cm. When it is under a load of 125 N, its total length is 20.5 cm. Calculate (a) its constant (k) (b) the load that makes it 25.0cm long
4. 4. Example : A linear spring has a length of 35.0 cm when under a load of 225 N and a length of 43.0 cm when under a load of 545 N. Find (a) its constant, and (b) its free (no load) length 225 N 35.0 cm 43.0 cm 545 N ΔF = k Δx
5. 5. ΔF = k Δx Solution: a) 545-225 = k (0.43-0.35) K= 320/0.08= 4000 N/m b) 545-0 = (40) (0.43-x0) x0= 0.43- (545/4000)= 0.2937 m x0= 29.37 cm
6. 6. The Linear Spring Formula: Note that the formula ΔF = k Δx is a relation between the applied force (to the spring) and the change in the spring's length. The spring force ( Fs ) is always opposite to the applied force. As the following figures indicate, when Fapplied is to the right, Δx is positive, Fs pulls to the left and is negative, and when Fapplied is to the left, Δx is negative , Fs pushes to the right and is positive.
7. 7. When ( x ) is positive, Fs is negative and vise versa. This fact is reflected by the ( - ) sign in the formula. formula for a linear spring Fs = - k x force that the spring exerts Fs is not the applied force
8. 8. Simple Harmonic Motion: If mass M performs a uniform circular motion in a vertical plane, its shadow on the x-axis performs a back-and-forth motion that is called simple harmonic motion. To understand the following figure, visualize that mass M moves slowly and counterclockwise along the circle (of radius A), and at different positions, picture its shadow on the floor. The angular position of mass M on the circle is determined by θ. Corresponding to every θ there is a shadow position measured by x from C to H. It is possible to relate x to θ. Since θ = ωt ; therefore, x can be related to ωt.
9. 9. ω = θ/t → θ = ωt
10. 10. The graph of x versus θ Maximum= " Amplitude " of oscillations
11. 11. Example : A bicycle wheel of radius 30.0 cm is spinning at a constant angular speed of 180 rpm in a vertical plane. Find a) its angular speed in rd/s. The shadow of a bump on its edge performs a oscillatory motion on the floor. (b) write the equation of the oscillations of the shadow knowing that the shadow is at its maximum at t = 0. (c) determine the distance of the shadow from the equilibrium position at t = 1.77 seconds.
12. 12. Solution: (a) ω = 180 (rev / min) ( 6.28rd / rev)( min / 60 sec ) = 18.8 rd /s (b) ω = 18.8 rd /s A = 30.0 cm x = A cos(ωt) x = (30.0 cm) cos (18.8t ) (c) t = 1.77sec x = (30.0cm) cos(18.8*1.77 rd ) = -8.56cm t in seconds
13. 13. Example : The equation of oscillations of a mass on a spring is given by x = 3.23 cos( 12.56t ) where x is in (cm) and t in seconds. Find its (a) Amplitude (b) angular speed (c) frequency and period of oscillations (d) its position at t = 0.112s Solution: x = A cos(ωt) (a) A = 3.23 cm (b) ω = 12.56 rd/s (c) ω =2πf f = ω / 2π f = 2 Hz T = 1/f T = 0.500 s (d) x = 3.23 cos( 12.56*0.112 rd) = 0.528 cm
14. 14. The Mass-Spring System a spring that is not loaded the same spring but loaded and stretched a distance ( - h ) loaded spring stretched further a distance ( -A ) and released the attached mass M oscillates up and down to (+A) and (-A) above and below the equilibrium level.
15. 15. angular speed of oscillations of mass spring
16. 16. Example : A 102-gram mass hung from a weak spring has stretched it by 3.00 cm. Let g = 9.81m/s2 and calculate (a) the load on the spring (b) the spring constant in N/m If the mass-spring system is initially in static equilibrium and motionless, and the mass is pushed up by +2.00 cm and released, calculate its (c) angular speed (d) Frequency (e) Period (f) the amplitude of oscillations (g) the equation of motion of such oscillations.
17. 17. Solution: (a) w = Mg w = (0.102kg)(9.81 m/s2) = 1.00N (b) ΔF = k Δx k = (1.00N) /( 0.0300m) = 33.3 N/m (c) ω = SQRT( k / M ) = SQRT [( 33.3 N/m ) / (0.102 kg)] = 18.1 rd/s (d) f = ω / (2π ) f = 2.88 Hz (e) T = 1 / f T = 0.347s ( f ) The 2.00 cm that the mass is pushed up above its equilibrium level, initially, becomes its amplitude. A = +2.00cm. (g) Knowing the constants A = 2.00cm and ω = 18.1 rd/s, the equation of motion becomes: x = 2.00 cm cos(18.1t ) In this equation, if we plug t = 0, we get X = +2.00cm. This is correct because at t = 0, the mass is released from X = +2.00cm.
18. 18. Example : The graph of x ( the distance from the equilibrium position ) versus time ( t ) for the oscillations of a mass-spring system is given below. For such oscillations, find (a) the amplitude (b) the period (c) the frequency (d) the angular speed (frequency) (e) the spring constant ( k ) if the mass of the object is 250 grams (f) the equation of motion for the oscillations
19. 19. Solution: (a) A = 2.00cm (b) T = 2 (0.125s) = 0.250 s (c) f = 1 / T f = 4.00 Hz (d) ω = 2π f ω = 2π (4.00/s) = 25.1 rd/s (e) ω = (k/M)(1/2) → ω2 = (k/M) → k = Mω2 k = (0.250kg)(25.1 rd/s)2 k = 158 N/m (f) x = A sin (ωt) x = (2.00cm)sin ( 25.12t ) The given graph is a sine function. Note that at t = 0, X = 0, according to the given graph. It is a sine function that is zero at t =0. and not a cosine function.
20. 20. Linear Velocity and Acceleration in Simple Harmonic Motion at x = +A or –A: force, and magnitude are maximum at x = 0: zero acceleration (because the spring is neither stretched or compressed, F = 0)
21. 21. Example : The equation of motion of a 22-kg log oscillating on ocean surface is x = 1.2 sin (3.14t) where x is in meters and t in seconds. Determine its, amplitude, angular speed (frequency), frequency, period, maximum speed, maximum acceleration (magnitude), and its position at to t = 0.19 s.
22. 22. Solution: A = 1.2 m ω = 3.14 rd/s f = ω/(2π) = 0.50 s-1 (Hz) T = 1/ f = 2.0 s |Vmax| = Aω |Vmax| = (1.2m)(3.14 rd/s) = 3.8 m/s (occurs at the middle) |amax| = Aω2 |amax| = (1.2m)(3.14rd/s)2 = 11.8 m/s2 Using the given equation, substituting for t, and putting the calculator in "Radians Mode," we get: x = 1.2 sin [ 3.14 (0.19)rd ] = 0.67 m
23. 23. Homework: Problems 1, 2( a, b), 4, 5, 9