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Simple Harmonic Motion

Simple Harmonic Motion. Adapted from Sautter 2003 – www.slideshare.net. 200 grams. Vibrating Tuning fork. A weight on a spring. A boy on a swing. Some Common Examples of Simple Harmonic Motion. Simple Harmonic Motion.

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Simple Harmonic Motion

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  1. Simple Harmonic Motion Adapted from Sautter 2003 – www.slideshare.net

  2. 200 grams Vibrating Tuning fork A weight on a spring A boy on a swing Some Common Examples of Simple Harmonic Motion

  3. 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.

  4. 200 grams 400 grams 600 grams Hooke's Law F = - k x \ Elongation of spring FORCE(N) Slope = spring constant ELONGATION (M)

  5. 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.

  6. Displacement = +max Velocity = 0 Acceleration = - max Kinetic Energy = 0 Net Force = - max Top of cycle Displacement = 0 Velocity = max Acceleration = 0 Kinetic Energy = max Net Force = 0 Mid cycle Displacement = - max Velocity = 0 Acceleration = + max Kinetic Energy = 0 Net Force = + max Bottom of cycle SHM in a Wt - Spring System CLICK HERE

  7. 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

  8. Correlating the Reference Circle, the Vertical View of Displacement & Simple Harmonic Motion Displacement Vector of Circular motion & Displacement in SHM

  9. 200 grams 200 grams 200 grams 200 grams SHM Displacement & the Reference Circle y = +max y = 0 y = 0 y = -max Displacement vector on Reference Circle Vertical View Simple Harmonic Position

  10. y = +max 900 y = 0 1800 y = 0 00 y = -max 2700 Note that the vertical view of the displacement vector is 0 at 00, 100 % upward at 900, 0 at 1800, 100 % downward at 2700 and finally 0 again at 3600 What trig function is 0 at 00, 1.0 (100%) at 900, 0 at 1800, -1.0 (100% and pointing down) at 2700, and 0 again at 3600 Displacement Equation for SHM The SINE y = Amp x sin θ Displacement vector on Reference Circle Vertical View

  11. Circular Motion with Velocity Vector The velocity vector is always tangent to the circular path

  12. 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.

  13. Correlating the Reference Circle, the Vertical View of Velocity & Simple Harmonic Motion Velocity Vector of Circular motion & Velocity in SHM

  14. 200 grams 200 grams 200 grams 200 grams SHM Velocity & the Reference Circle V = 0 V = + max V = -max V = 0 Velocity vector on Reference Circle Vertical View Simple Harmonic Position

  15. V = 0 900 V = + max 00 V = -max 1800 V = 0 2700 Note that the vertical view of the velocity vector is 100 % upward at 00, 0 at 900, 100% downward at 1800, 0 at 2700 and finally 100% again at 3600 What trig function is 1.0 (100%) at 00, 0 at 900, 1.0 (100%) at 1800, 0 at 2700, and 1.0 again at 3600 Velocity Equation for SHM Velocity vector on Reference Circle Vertical View The COSINE V = Vmax x cos θ

  16. Correlating the Reference Circle, the Vertical View of Acceleration & Simple Harmonic Motion Acceleration Vector of Circular motion & Acceleration in SHM

  17. 200 grams 200 grams 200 grams 200 grams SHM Acceleration & the Reference Circle a = -max a = 0 a = 0 a = +max Acceleration vector on Reference Circle Vertical View Simple Harmonic Position

  18. a = -max, 900 a = 0 00 a = 0 1800 a = +max, 2700 Acceleration Equation for SHM Note that the vertical view of the acceleration vector is 0 at 00, 100 % downward at 900, 0 at 1800, 100 % upward at 2700 and finally 0 again at 3600 What trig function is 0 at 00, -1.0 (100%) at 900, 0 at 1800, +1.0 (100% and pointing down) at 2700, and 0 again at 3600 The - SINE a = amax x ( -sin θ) Acceleration vector on Reference Circle Vertical View

  19. 200 grams 200 grams 200 grams 200 grams 200 grams Simple Harmonic Motion 90 o 0 o 180 o 360 o 270 o Y = Amplitude x Sin  Acc. = Acc. max x (-Sin ) V = Velocity max x Cos 

  20. Simple Harmonic Motion • = 2  f t V = 2  f R V = 2  R / T ac = 4 2 f 2 R ac = 4 2 R/ T2 • Recall the following relationships pertaining to circular motion. R = the radius of the reference circle.

  21. Expanded Displacement Equations The amplitude (A) = the radius of the reference circle (R) y = A x sin θ y = A x sin o t y = A x sin 2  f t Expanded Velocity Equations V = Vmax 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

  22. Expanded Acceleration Equations The amplitude (A) = the radius of the reference circle (R) a = amax x ( -sin θ) a = V2 / A x ( -sin θ) a = 4 2 f 2 A x ( -sin θ) a = 4 2 A/ T2 x ( -sin θ) a = amax x ( -sin 2  f t )

  23. π T = 2 m / k Deriving the Equation for Period of aWeight – Spring System • Fspring = Facceleration of mass • -kx = ma • The elongation of the spring equals the amplitude (A) of the vibration. • a = 4 2 A/ T2 x ( -sin θ) • -kA =m 4 2 A/ T2 x ( -sin θ) • The amplitude of vibration is reached when θ = 90 degrees. Sine of 900 = 1.0 • Substituting 1.0 for sin θ and rearranging: • T2 = 4 2 m A/ kA = 4 2 m/k • Solving for T gives:

  24. T = 2 m / k π Small masses vibrate with shorter periods Large masses vibrate with longer periods

  25. T = 2 m / k π Springs with larger constants vibrate with shorter periods Springs with smaller constants vibrate with longer periods

  26. The Simple Pendulum T = 2l / g Fc (mg cos θ) θ T (tension) centripetal force W θ PEmax = mg h KE =0 Fr Vector Diagram h Fr Restoring Force Fcentrifugal mg (weight) KEmax = ½ mv2 PE =0

  27. π T = 2 L / g Deriving the Equation for Period of aSimple Pendulum • 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 = mac • Centripetal acceleration is given by: a = 4 2 R/ T2 • mg cos θ = m 4 2 R/ T2 • The radius of the circle is the pendulum length (L). Rearranging the equation gives: • T2 = 4 2 L/ g cos θ, for small angles cos θ  1.0 and: T2 = 4 2 L/ g , taking square roots gives:

  28. T = 2 L / g π SHORT PENDULUMS HAVE A SHORT PERIOD OF OSCILLATION LONG PENDULUMS HAVE A LONG PERIOD OF OSCILLATION

  29. T = 2 L / g π STRONGER GRAVITY FIELDS RESULT IN SHORTER PERIODS OF OSCILLATION GRAVITY ON EARTH 9.8 M/S2 GRAVITY ON MOON 1.6 M/S2

  30. T = 2 m / k π T = 2 L / g π Weight - Spring System Pendulum

  31. 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 00 and 1800 (the midpoint of the cycle) • V = 2  f A x cos θ • V = 2  10 x 5.0 x cos 00 = 100  (1.0) = 313 cm/sec • (b) the maximum acceleration occurs at 900 and 2700 (the endpoints of the cycle) • a = 4 2 f 2 A x ( -sin θ) • a = 4 2 (10)2 x 10 x ( -sin 900) = 4000 2 ( -1.0) • a = 39,500 cm/s2 or 39.5 m/s2

  32. A y θ Reference circle Solving SHM ProblemsAn 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. y = 5.0 cm, A = 10 cm Sin θ = 5.0 / 10 = . 50 Sin –1 (0. 50) = 300 • (a) V = 2  A/ T x cos θ, V = 2  10 / 2.0 x cos 300 • V = 10  x 0.866 = 27.2 cm/s • (b) a = 4 2 A/ T2 x ( -sin θ), a = 4 2 10/ 22 x ( -sin 300) • a = 4 2 10/ 2.0 2 x ( -0.50) = 5.0 2 = 49.3 cm/s2

  33. T = 2 m / k π 5 lbs Solving SHM ProblemsA 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. F = - kx • (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)

  34. T = 2 m / k π 30 Nts Solving SHM ProblemsA 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 = 4π2 m / T2 = 4 π2 (30/9.8) / 0.302 = 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

  35. θ Solving SHM ProblemsA pendulum 1.00 meters long oscillates at 30 times a minute. What is the value of gravity ? T = 2l / g • 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 / T2 • g = 4 2 x 1.00 / 2.02 = 9.87 m/s2

  36. 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 !

  37. 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 Click here for answers • 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 100 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

  38. The End

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