Your SlideShare is downloading. ×
0
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×
Saving this for later? Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.
Text the download link to your phone
Standard text messaging rates apply

Simple Harmonic Motion

50,774

Published on

Describes simple harmonic motion of weight- spring systems and pendulums. …

Describes simple harmonic motion of weight- spring systems and pendulums.
**More good stuff available at:
www.wsautter.com
and
http://www.youtube.com/results?search_query=wnsautter&aq=f

Published in: Education, Technology, Business
1 Comment
3 Likes
Statistics
Notes
  • Write a comment...ALSO
    I have written six books
    'Sticks - A Golfer’s Tale' - A hacker’s dream comes true.
    'Fish Farm' - Revenge of the old people.
    'Coach' - A mystery in Old school football town in a rural, bigoted, fifties town.
    'The Three Dollar Phoenix' - A mystery set in Newark, New Jersey in the 1970s.
    'The Divine Comedy MMIX' - A humorous play about Jesus returning.
    'The Blood of Judas' - A horror story of revenge set in Nazi Germany.
    All are available at www.smashwords.com
    I have video trailers for 'Coach', 'Fish Farm' and 'The Blood of Judas' at:
    http://www.youtube.com/watch?v=xXSD5Kz-fDY
    http://www.youtube.com/watch?v=a9PTRb14ldc
    http://www.youtube.com/watch?v=ToPp9k9Oq-o
    http://www.youtube.com/watch?v=3eBhMZbsP-I
    Please take a look. Thanks.
    Walt Sautter - wsautter@optonline.net
       Reply 
    Are you sure you want to  Yes  No
    Your message goes here
No Downloads
Views
Total Views
50,774
On Slideshare
0
From Embeds
0
Number of Embeds
1
Actions
Shares
0
Downloads
406
Comments
1
Likes
3
Embeds 0
No embeds

Report content
Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
No notes for slide

Transcript

  • 1. Simple Harmonic Motion Copyright Sautter 2003
  • 2. The next slide is a quick promo for my books after which the presentation will begin Thanks for your patience! Walt S. [email_address] More stuff at: www. wsautter .com
  • 3. Books available at: www. wsautter .com www.smashwords.com www.amazon.com www.bibliotastic.com www.goodreads.com Walt’s Books for Free!
  • 4. Some Common Examples of Simple Harmonic Motion Vibrating Tuning fork A weight on a spring A boy on a swing 200 grams
  • 5. Simple Harmonic Motion
    • Simple harmonic motion (SHM) is a repeated motion of a particular frequency and period.
    • The force causing the motion is in direct relationship to the displacement of the body. (Hooke’s Law)
    • The displacement, velocity, acceleration and force characteristics are specific a various points in the cycle for SHM.
    • SHM can be understood in terms of the displacement, velocity, acceleration and force vectors related to circular motion.
    • Recall that the displacement vector for circular motion is the radius of the circular path. The velocity vector is tangent to the circular path and the acceleration vector always points towards the center of the circle.
  • 6. Hooke's Law F = - k x F O R C E (N) ELONGATION (M) Slope = spring constant Elongation of spring 400 grams 200 grams 600 grams
  • 7. Simple Harmonic Motion
    • SHM motion can be represented as a vertical view of circular motion. Using this concept, we can see the variations in the vector lengths and directions for displacement, velocity and accelerations as those values for SHM.
    • Use the following slide showing a mass on a spring, vibrating in SHM to examine the variations in these three vectors as the reference circle rotates.
    • See if you can decide which trig functions (sine, cosine or tangent) govern each to the three vectors in SHM
  • 8. Displacement = +max Velocity = 0 Acceleration = - max Kinetic Energy = 0 Net Force = - max Displacement = 0 Velocity = max Acceleration = 0 Kinetic Energy = max Net Force = 0 Top of cycle Mid cycle Bottom of cycle Displacement = - max Velocity = 0 Acceleration = + max Kinetic Energy = 0 Net Force = + max SHM in a Wt - Spring System CLICK HERE
  • 9. Velocity & Acceleration Vectors in Circular Motion The velocity vector (black) is always directed tangentially to the circular path. The acceleration vector (red) is always directed toward the center of the circular path
  • 10. Correlating the Reference Circle, the Vertical View of Displacement & Simple Harmonic Motion Displacement Vector of Circular motion & Displacement in SHM Acceleration Vector
  • 11. Displacement vector on Reference Circle SHM Displacement & the Reference Circle Vertical View Simple Harmonic Position y = +max y = 0 y = -max y = 0 200 grams 200 grams 200 grams 200 grams
  • 12. Displacement vector on Reference Circle Vertical View Note that the vertical view of the displacement vector is 0 at 0 0 , 100 % upward at 90 0 , 0 at 180 0 , 100 % downward at 270 0 and finally 0 again at 360 0 What trig function is 0 at 0 0 , 1.0 (100%) at 90 0 , 0 at 180 0 , -1.0 (100% and pointing down) at 270 0 , and 0 again at 360 0 The SINE y = Amp x sin θ Displacement Equation for SHM y = 0 0 0 y = -max 270 0 y = 0 180 0 y = +max 90 0
  • 13. Circular Motion with Velocity Vector The velocity vector is always tangent to the circular path
  • 14. Vertical Component of Circular Motion The reference circle is turned sideways and viewed vertically. This shows the velocity vector of a body in Simple Harmonic Motion.
  • 15. Velocity Vector of Circular motion & Velocity in SHM Correlating the Reference Circle, the Vertical View of Velocity & Simple Harmonic Motion
  • 16. Velocity vector on Reference Circle SHM Velocity & the Reference Circle Vertical View Simple Harmonic Position V = 0 V = + max V = 0 V = -max 200 grams 200 grams 200 grams 200 grams
  • 17. Note that the vertical view of the velocity vector is 100 % upward at 0 0 , 0 at 90 0 , 100% downward at 180 0 , 0 at 270 0 and finally 100% again at 360 0 What trig function is 1.0 (100%) at 0 0 , 0 at 90 0 , 1.0 (100%) at 180 0 , 0 at 270 0 , and 1.0 again at 360 0 Velocity vector on Reference Circle Vertical View The COSINE V = V max x cos θ Velocity Equation for SHM V = 0 90 0 V = + max 0 0 V = 0 270 0 V = -max 180 0
  • 18. Correlating the Reference Circle, the Vertical View of Acceleration & Simple Harmonic Motion Acceleration Vector of Circular motion & Acceleration in SHM
  • 19. Acceleration vector on Reference Circle SHM Acceleration & the Reference Circle Vertical View Simple Harmonic Position a = -max a = 0 a = +max a = 0 200 grams 200 grams 200 grams 200 grams
  • 20. Acceleration vector on Reference Circle Vertical View Note that the vertical view of the acceleration vector is 0 at 0 0 , 100 % downward at 90 0 , 0 at 180 0 , 100 % upward at 270 0 and finally 0 again at 360 0 What trig function is 0 at 0 0 , -1.0 (100%) at 90 0 , 0 at 180 0 , +1.0 (100% and pointing down) at 270 0 , and 0 again at 360 0 The - SINE a = a max x ( -sin θ) Acceleration Equation for SHM a = -max, 90 0 a = 0 0 0 a = +max, 270 0 a = 0 180 0
  • 21. Simple Harmonic Motion 0 o 90 o 180 o 270 o 360 o V = Velocity max x Cos  Acc. = Acc. max x (-Sin  ) Y = Amplitude x Sin  200 grams 200 grams 200 grams 200 grams 200 grams
  • 22. y t Y = Amp x Sin  Amp dy/dt t dy/dt = V max Cos  d 2 y/dt 2 t d 2 y/dt 2 = A max (-Sin  ) Derivative Functions Displacement Velocity Acceleration
  • 23. Simple Harmonic Motion
    • Recall the following relationships pertaining to circular motion. R = the radius of the reference circle.
    • =  o t
    •  = 2  f
    •  = 2  / T
    • V linear =  R
    • a centripetal = V 2 / r
    COMBINING THESE EQUATIONS WE GET:
    • = 2  f t
    • V = 2  f R
    • V = 2  R / T
    • a c = 4  2 f 2 R
    • a c = 4  2 R / T 2
  • 24. Expanded Displacement Equations y = A x sin θ y = A x sin  o t y = A x sin 2  f t Expanded Velocity Equations V = V max x cos θ V = 2  f A x cos θ V = 2  f A x cos  o t V = 2  f A x cos 2  f t V = 2  A/ T x cos 2  f t The amplitude (A) = the radius of the reference circle (R)
  • 25. Expanded Acceleration Equations The amplitude (A) = the radius of the reference circle (R) a = a max x ( -sin θ) a = V 2 / A x ( -sin θ) a = 4  2 f 2 A x ( -sin θ) a = 4  2 A / T 2 x ( -sin θ) a = a max x ( -sin 2  f t )
  • 26. Deriving the Equation for Period of a Weight – Spring System
    • F spring = F acceleration of mass
    • -kx = ma
    • The elongation of the spring equals the amplitude (A) of the vibration.
    • a = 4  2 A / T 2 x ( -sin θ)
    • -kA =m 4  2 A / T 2 x ( -sin θ)
    • The amplitude of vibration is reached when θ = 90 degrees. Sine of 90 0 = 1.0
    • Substituting 1.0 for sin θ and rearranging:
    • T 2 = 4  2 m A / kA = 4  2 m/k
    • Solving for T gives:
    T = 2 m / k π
  • 27. Small masses vibrate with shorter periods Large masses vibrate with longer periods T = 2 m / k π
  • 28. Springs with smaller constants vibrate with longer periods Springs with larger constants vibrate with shorter periods T = 2 m / k π
  • 29. mg (weight) T (tension ) centripetal force F r Restoring Force W F c (mg cos θ) F r  h PE max = mg  h KE =0 KE max = ½ mv 2 PE =0 T = 2  l / g The Simple Pendulum Vector Diagram θ θ F centrifugal
  • 30.
    • The pendulum experiences centripetal acceleration as it swing in an arc. The component of the weight causing tension in the string which supplies the centripetal force is given by m x g cos θ.
    • mg cos θ = ma c
    • Centripetal acceleration is given by: a = 4  2 R / T 2
    • mg cos θ = m 4  2 R / T 2
    • The radius of the circle is the pendulum length (L). Rearranging the equation gives:
    • T 2 = 4  2 L / g cos θ, for small angles cos θ  1.0 and: T 2 = 4  2 L / g , taking square roots gives:
    Deriving the Equation for Period of a Simple Pendulum T = 2 L / g π
  • 31. SHORT PENDULUMS HAVE A SHORT PERIOD OF OSCILLATION LONG PENDULUMS HAVE A LONG PERIOD OF OSCILLATION T = 2 L / g π
  • 32. GRAVITY ON EARTH 9.8 M/S 2 GRAVITY ON MOON 1.6 M/S 2 STRONGER GRAVITY FIELDS RESULT IN SHORTER PERIODS OF OSCILLATION T = 2 L / g π
  • 33. Pendulum Weight - Spring System T = 2 m / k π T = 2 L / g π
  • 34. Torsional (Twist) Pendulums act because of Hooke’s Law applied in rotational form τ = - k θ I (Moment of Inertia) replaces mass in the Weight- spring system equation The Torsional Pendulum T = 2 I / k π T = 2 m / k π
  • 35. Solving SHM Problems
    • What is the maximum velocity of an object in SHM with a amplitude of 5.0 cm and a frequency of 10 hertz? What is its maximum acceleration ?
    • Solution: (a) the maximum velocity occurs at 0 0 and 180 0 (the midpoint of the cycle)
    • V = 2  f A x cos θ
    • V = 2  10 x 5.0 x cos 0 0 = 100  (1.0) = 313 cm/sec
    • (b) the maximum acceleration occurs at 90 0 and 270 0 (the endpoints of the cycle)
    • a = 4  2 f 2 A x ( -sin θ)
    • a = 4  2 (10) 2 x 10 x ( -sin 90 0 ) = 4000  2 ( -1.0 )
    • a = 39,500 cm/s 2 or 39.5 m/s 2
  • 36. Solving SHM Problems An object vibrates with an amplitude of 10 cm and a period of 2.0 seconds. Find the velocity and acceleration when the displacement is 5.0 cm.
    • (a) V = 2  A/ T x cos θ, V = 2  10 / 2.0 x cos 30 0
    • V = 10  x 0.866 = 27.2 cm/s
    • (b) a = 4  2 A / T 2 x ( -sin θ), a = 4  2 10 / 2 2 x ( -sin 30 0 )
    • a = 4  2 10 / 2.0 2 x ( -0.50 ) = 5.0  2 = 49.3 cm/s 2
    y = 5.0 cm, A = 10 cm Sin θ = 5.0 / 10 = . 50 Sin –1 (0. 50) = 30 0 A y θ Reference circle
  • 37. Solving SHM Problems A spring with a constant of 40 lbs/ft has a 5 lb object suspended from it. Find its elongation and period of vibration when set into motion.
    • (a) F = - kx, x = -F /k
    • x = - (-5 lbs/ 40 lbs/ft) = 0.125 ft or 1.5 inches (note that the negative – 5 lbs means that the weight acts downward)
    • (b) T = 2 π (( 5/ 32) / 40) 1/2 = 0.393 seconds ( note that the 5/32 converts weight in pounds to mass in slugs)
    F = - kx 5 lb s T = 2 m / k π
  • 38. Solving SHM Problems A spring has a 0.30 second period of vibration when a 30 N weight is hung from it. How much will it stretch when a 50 N weight is hung from it ?
    • To find the the stretch of the spring we must use Hooke’s Law and know the spring constant.
    • Using T = 2 π (m/k) 1/2 , we can solve for k as:
    • k = 4 π 2 m / T 2 = 4 π 2 (30/9.8) / 0.30 2 = 1343 N/ m (note 30 / 9,8 gives the mass of the object)
    • Now using Hooke’s Law : F = - kx, x = -F / k
    • X = 50 / 1343 = 0.0372 m or 3.72 cm
    F = - kx 30 Nts T = 2 m / k π
  • 39. Solving SHM Problems A pendulum 1.00 meters long oscillates at 30 times a minute. What is the value of gravity ?
    • 30 oscillations per minute = 30 / 60 = 0.50 oscillations per second
    • f = 1/ T, f = 1 / 0.50 = 2.0 sec
    • T = 2  (l / g) 1/2 , g = 4  2 l / T 2
    • g = 4  2 x 1.00 / 2.0 2 = 9.87 m/s 2
    T = 2  l / g θ
  • 40. Solving SHM Problems A torsion pendulum consists of a 2.0 Kg solid disk of 15 cm radius. When a 1 N-m torque is applied it displaces 15 0 . What is its frequency of vibration?
    • The moment of inertia of a disk is given by: I = ½ mr 2
    • I = ½ (2.0) (.15) 2 = 0.0225 Kg-m 2 (note: 15 cm = .15 m)
    • τ = - k θ, k = τ / θ, (15 0 in radians = 0.262 rad)
    • k = 1 / 0.262 = 3.82 N-m / radian
    • T = 2 π (0.0225 / 0.262) 1/2 = 0.482 seconds
    • f = 1 / T = 1 / 0.482 = 2.07 hertz
    τ = - k θ T = 2 I / k π
  • 41. Now it's time for you to try some problems on your own ! The problems are similar to the ones which have been solved so look back and review the appropriate problem if you get stuck !
  • 42. A spring 60 mm long is stretched by 5.0 mm when a 200 gram mass is suspended from it. What is its spring constant in N/m ? (A) 1.6 (B) 40 (C) 196 (D) 392
    • An object in SHM has an amplitude of 12 mm and a period of 0.40 seconds.
    • What is its maximum velocity in cm per second ?
    • 3.0 (B) 19 (C) 38 (D) 43
    A swing moves back and forth every 4.0 seconds. How long is the swing in meters? (A) 2.5 (B) 4.0 (C) 6.2 (D) none of these A 24 kg ball with a radius of 20 cm is suspended from a wire. It is Rotated through 10 0 when a torque of 0.50 N-m is applied. Find the period. (A) 2.30 sec (B) 1.15 sec (C) 5.3 sec (D) 1.11 sec In order to change the frequency of a mass – spring system by a factor of 2 the mass must be multiplied by a factor of (A) 4 (B) 2 (C) ½ (D) ¼ Click here for answers
  • 43. The End

×