Gelombang MekanikGelombang mekanik merupakan gelombang yang membutuhkan medium untuk berpindah tempat. Gelombang laut, gelombang tali atau gelombang bunyi termasuk dalam gelombang mekanik. Kita dapat menyaksikan gulungan gelombang laut karena gelombang menggunakan laut sebagai perantara. Kita bisa mendengarkan musik karena gelombang bunyi merambat melalui udara hingga sampai ke telinga kita. Tanpa udara kita tidak akan mendengarkan bunyi. Dalam hal ini udara berperan sebagai medium perambatan bagi gelombang bunyi.
Mechanical wave is a wave that requires a medium to adjourn. Ocean waves, waves or sound waves strap included in the wave mechanics. We can watch the ocean waves because the waves roll using the sea as an intermediary. We can listen to music because the sound waves propagate through the air until he came to our ears. Without air we're not going to listen to the sound. In this case the air acts as a medium for sound wave propagation.
One. Wave mechanics (formula) is the wave propagation phenomena of energy. Mechanical waves are waves which require a medium to propagate. A = amplitude waves (m) = = = wavelength (m) v = propagation of the wave (m / s) = = = frequency angles (rad / s) k = = number of waves (m-1) x = distance between point of the origin (m) Particle Acceleration Particle Velocity Phase Angle Phase Waves Waves Waves Walking is the wave amplitude is fixed. Stationary wave is the wave amplitude is changed. Stationary wave Bound Edge (Node) Free Edge stationary waves (Stomach) Location of the node to - (n +1) from the end location of the node bound to - (n +1) from the free end to the Location of the belly - (n +1) from the tip bound Location of the node to - (n +1) from the tip of a transverse wave propagation Fast bound in string v = wave propagation in the string (m / s) voltage wire F = force (N) = linear strings of mass (kg / m) m = mass String (kg) l = length of string (m) = density of string material (kg/m3) cross-sectional area A = string (m2) The frequency of sound produced by strings f1: f 2: F3: ... = 1: 2: 3: ... © Aidia Propitious
2. Organa Pipe Pipe Closed Open Organa distance between two adjacent stomach: The distance between the stomach and adjacent node: Basic Tone: Tone basis: F0: f1: f2: ... = 1: 2: 3: ... F0: f1: f2: ... = 1: 3: 5: Doppler Effect ... VP = + (listener approached source) listener frequency fp = vp = 0 (silent listener) fs = frequency of the source VP = - (listener away from the source), v = velocity of sound in air v = - (a source close to listener) listener velocity vp = vs = 0 (source rest) source velocity vs = vs = + (source away from the listener) Intensity noise sound P = power (watts) A = = broad field of the ball (m2) R = distance of a point to the source sound (m) f = frequency (Hz) = the density of the material (kg/m3), A = amplitude (m) v = wave propagation speed (m / s) TI Sound Intensity Level = Sound Intensity Level (decibel = dB) I0 = Intensity threshold = 10-16 100 Hz watts/cm2 Limits I and IT that can be heard at 100 Hz: 10-16 I 10-4 watts/cm2 or 0 IT 120dB (Example Problem Wave Mechanics) 1. End of a rope vibrated harmonic with a period of 0.5 s and an amplitude of 6 cm. These vibrations propagate to the right along the ropes with fast velocity 200 cm / s. Define: a. General equation of wave b. Deviation, velocity, and acceleration of particles in P which is 27.5 cm from the end of the rope is vibrated when the vibrating tip vibrating and shaking 0.2 s c. Phase and phase angle of particles in the P now has vibrating tip vibrating d. 0.2 s Antra phase difference of two particles along the rope, a distance of 25 cm © Aidia Propitious 2
3. Answer: a. T = 0.5 s; A = 6 cm; v = 200 cm / s; gel. propagating to the right equation of the wave: b. x = 27.5 cm; t = 0.2 s deviation: Speed: Acceleration: c. Phase angle:; Phase: d. x = 25 cm; Different phases: 2. The equation of a transverse wave is propagating along the wire is given by: Compute: a. Propagation of the wave b. Maximum velocity of a particle in the wire Answer: a. Fast wave propagation: A = 2 mm; k = 20 m-1; ω = 600 s-1 b. Maximum speed: 3. A traveling wave on a wire is given by the equation: Where x and y in cm and t in seconds. Specify: © Aidia Propitious 3
4. a. B. The direction of wave propagation Amplitude, frequency, wavelength and propagation of a wave of c. Maximum acceleration of a particle in the rope Answer: a. The wave propagation direction: to the left because the phase angle (+) b. Amplitude: A = 2 cm, frequency and wavelength: Fast wave velocity: c. Acceleration: Maximum Acceleration: 4. A sinusoidal wave with a frequency of 500 Hz has a propagation of 350 m / s. a. What is the distance between two points separating the different phases of π / 3 rad? b. What is the phase difference at a different particle 1 ms time? Answer: f = 500 Hz; v = 350 m / s a. The distance separating two points: = π / 3 b. Phase difference of a particle: t = t2 - t1 = 1 ms = 1 x 10-3 s 5. A wire whose length is 100 cm stretched horizontally. Harmonic one end vibrated up and down with frequency 1 / 8 Hz and an amplitude of 16 cm, while the other end tied. The harmonic vibrations propagate to the right along the wire with the propagation of 4.5 cm / s. Determine the location of the node to-4 and-3 from the stomach into the origin of vibration! Answer: L = 100 cm, f = 1 / 8 Hz; A = 16 cm, v = 4.5 cm / s © Aidia Propitious 4
5. Node number - 4 n + 1 = 4, n = 3 Location of the node to - four from the origin = L - x4 = 100-54 = 46 cm into the stomach - 3 n + 1 = 3, n = 2 Location of Abdominal - 3 from the origin = 100-45 = 55 cm 6. One end of a rope whose length is 115 cm harmonic vibrated up and down, while the other end free to move. a. What is the wavelength of the propagating on the string if the stomach to-3 is 15 cm from the origin of vibration? b. Where is the second node from the origin of vibration measured? Answer: a. x3 = 15; to-3 n + 1 = 3, n = 2 b. to-2 n + 1 = 2, n = 1 Location of the second node from the origin Vibrating = L - x2 = 115-11.25 = 103.75 cm 7. Vibrations of a spring whose length is 60 cm and tied at both ends in accordance with: Where the x and y in cm and t in s. a. What is the maximum deviation of a point at x = 5 cm? b. What is the location of nodes along the spring? c. What is the velocity of particles at x = 7.5 cm at t = 0.25 s? Answer: a. The maximum y value if the value of the maximum πt cos 96, ie 96 cos πt = a: b. Node has the deviation (y) = 0 © Aidia Propitious
6. So: c. Speed is derived from the deviation of x = 7.5 cm and t = 0.25 s 8. A piece of string tied to both ends vibrated. How many nodes and abdomen when the strings ring tones on the 5th? Answer: For the basic tone of the two ends of strings there are two knots clamped a stomach, and as applicable to the basic tone-5 was obtained by adding 5 knots on the basic tone, thus: So the tone of the 5th and 6th there are seven knots stomach. (Example Problem Frequency Sound) 1. Piano string whose length is 0.5 m and a mass of 10-2 kg tensed 200 N. Calculate: a. Transverse wave propagation in the string b. The frequency of the basic tone of the piano c. Overtone frequency to the one and two pianos Answer: L = 0.5 m, m = 10-2 kg, F = 200 N a. Propagation of the b. C. basic tone frequency Tone for his first and 2nd F0: f1: f2 = 1: 2: 3 © Aidia Propitious 6
7. engar in other cars before and after the two passed each other! Answer: Before passing: vp = + 14 m / s; vs = - 14 m / s; v = 334 m / s; fs = 640 After passing: vp = - 14 m / s; vs = + 14 m / s © Aidia Propitious 11