Copyright Sautter 2003 Vibrating Tuning fork A weight

Copyright Sautter 2003

Vibrating Tuning fork A weight on a spring 200 grams A boy on a swing

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.

Elongation of spring F O R C E (N) 200 grams 400 grams 600 Slope = spring constant ELONGATION (M) grams

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

Top of cycle Mid cycle Bottom of cycle CLICK HERE Displacement = +max Velocity = 0 Acceleration = - max Kinetic Energy = 0 Net Force = - max Displacement = 0 Velocity = max Acceleration = 0 Kinetic Energy = max Net Force = 0 Displacement = - max Velocity = 0 Acceleration = + max Kinetic Energy = 0 Net Force = + max

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

Displacement Vector of Circular motion & Displacement in SHM

y = +max 200 grams y=0 200 grams y = -max Displacement vector on Reference Circle Vertical View Simple Harmonic Position

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 y = +max 900 y=0 1800 y=0 00 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 y = -max 2700 Displacement vector on Reference Circle Vertical View The SINE y = Amp x sin θ

The velocity vector is always tangent to the circular path

The reference circle is turned sideways and viewed vertically. This shows the velocity vector of a body in Simple Harmonic Motion.

Velocity Vector of Circular motion & Velocity in SHM

V=0 V = -max 200 V = + max grams 200 grams V=0 Velocity vector on Reference Circle Vertical View Simple Harmonic Position

V=0 900 V = + max 00 V = -max 1800 Velocity vector on Reference Circle 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 V=0 2700 Vertical View The COSINE V = Vmax x cos θ

Acceleration Vector of Circular motion & Acceleration in SHM

a = -max a=0 200 grams a=0 200 grams a = +max Acceleration vector on Reference Circle Vertical View Simple Harmonic Position

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 a = -max, 900 a=0 1800 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 a = +max, 2700 Acceleration vector on Reference Circle Vertical View The - SINE a = amax x ( -sin θ)

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

Amp Y = Amp x Sin y Displacement t dy/dt = Vmax Cos Velocity dy/dt t d 2 y/dt 2 = Amax (-Sin ) Acceleration d 2 y/dt 2 t

Simple Harmonic Motion • Recall the following relationships pertaining to circular motion. R = the radius of the reference circle. q = o t =2 f =2 /T Vlinear = R acentripetal = V 2 / r COMBINING THESE EQUATIONS WE GET: q=2 ft V=2 f. R V=2 R/T ac = 4 2 f 2 R ac = 4 2 R / T 2

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 o t V = 2 f A x cos 2 f t V = 2 A/ T x cos 2 f t

Expanded Acceleration Equations The amplitude (A) = the radius of the reference circle (R) a = amax 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 = amax x ( -sin 2 f t )

Deriving the Equation for Period of a Weight – Spring System • Fspring = Facceleration 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) • -k. A =m 4 2 A / T 2 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: • T 2 = 4 2 m A / k. A = 4 2 m/k • Solving for T gives: π

Small masses vibrate with shorter periods Large masses vibrate with longer periods π

Springs with larger constants vibrate with shorter periods π Springs with smaller constants vibrate with longer periods

T = 2 l/g θ PEmax = mg h T W (tension) centripetal force KE =0 h KEmax = ½ mv 2 PE =0 Fr Restoring Force Fc θ (mg cos θ) Fr Vector Diagram Fcentrifugal mg (weight)

Deriving the Equation for Period of a Simple 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 / 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: π

SHORT PENDULUMS HAVE A SHORT PERIOD OF OSCILLATION π LONG PENDULUMS HAVE A LONG PERIOD OF OSCILLATION

STRONGER GRAVITY FIELDS RESULT IN SHORTER PERIODS OF OSCILLATION GRAVITY ON EARTH 9. 8 M/S 2 GRAVITY ON MOON 1. 6 M/S 2 π

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 π π

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/s 2 or 39. 5 m/s 2

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 y θ Reference circle • • 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 / T 2 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/s 2

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. F = - kx 5 lbs π • (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)

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 ? F = - kx 30 Nts π • To find 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 π2 m / T 2 = 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

Solving SHM Problems A pendulum 1. 00 meters long oscillates at 30 times a minute. What is the value of gravity ? θ 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 / T 2 • g = 4 2 x 1. 00 / 2. 02 = 9. 87 m/s 2

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 150. What is its frequency of vibration? τ=-k θ π • 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, (150 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

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 ? (A) 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