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GEJALA GELOMBANG

GEJALA GELOMBANG. A. Gelombang berjalan. Yo = Asin θ. Fase gel. Ψ p=(t/T-x/ λ )= θ p/2 π Sudut fase θ p=( ω t- kx ) Beda fase Δψ =( ψ b- ψ a)=-( xb-xa ) λ. V. Yo = Asin ω t Yp = A sin 2 π t/T Yo = A sin 2 πψ Yp = A sin θ p Yp = Asin 2 πψ p Yp = A sin2 π tp /T

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GEJALA GELOMBANG

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  1. GEJALA GELOMBANG A. Gelombangberjalan Yo= Asinθ Fase gel. Ψp=(t/T-x/λ)=θp/2π Sudutfaseθp=(ωt-kx) Beda faseΔψ=(ψb-ψa)=-(xb-xa) λ V Yo= Asinωt Yp= A sin 2πt/T Yo= A sin 2πψ Yp= A sin θp Yp= Asin 2πψp Yp= A sin2πtp/T Yp= A sin 2π(t-x/v) T Yp= A sin2π(t/T-x/vT) Yp= A sin2π(t/T-x/λ) Yp= A sin(2πt/T-2πx/λ) {2π/T=ωdan 2π/λ=k} Yp= A sin(ωt-kx) P O X PERSAMAAN UMUM: Titikasal gel merambat Keataskekanan ↓ ↓ Y= ± A sin (ωt kx) ↑ ↑ Titikasal gel merambat Kebawahkekiri Kecepatan V = ± ωA sin (ωt kx) Percepatan a = ± ω2A sin (ωt kx)

  2. B. Gelombangstasioner: a. ujungtetap ; y= 2A sin kxcosωt y= As cosωt As= 2A sin kx Letaksimpul; Xn+1=(2n X λ/4) Letakperut : Xn+1=(2n +1)λ/4 b. Ujung Bebas : y=2A coskx sin ωt y=As sin ωt As=2A coskx Letaksimpul; Xn+1=(2n +1) λ/4) Letakperut : Xn+1=(2n X λ/4) n= 0.1,2,3………

  3. GELOMBANG BUNYI • Frekuensibunyi : • Frekuensi audio manusia : 20 Hz - 20000Hz • Frekuensiinfrasonik <20 Hz • FrekuensiUltrasonik >20000Hz B. Medium Perambatan: 1. Gas→ 2. ZatCair→ 3. Zatpadat→ 4. Dawai/senar→ Keterangan: v=Cepatrambatbunyi(m/s) γ= Konstanta Laplace R= Tetapan gas umum T= Suhu M=Massa satu mol Gas B = modulus bulk zatcair 𝛒=Massa jenis F=Gaya 𝜇=massa per satuanpanjang, 𝜇=m/l

  4. C. Sumber-sumberbunyi: 1. Dawai/senar : a. Frekuensi nada dasar →fo=v/λ→l=1/2λ→λ=2l (Harmonik ke-1) fo= v =√F/𝜇 2l 2l b. Frekuensi nada atas pertama→f1→v/λ→l=λ (Harmonik ke-2) f1= v =√F/𝜇 l l C. Frekuensi nada atas kedua→f2→v/λ→l=(3/2)λ (Harmonik ke-3) f2 = 3v = 3√F/𝜇 2l 2l l=(1/2)λ P S S P P S S S l=λ P P P S S S S l=(3/2)λ d. fn=(n+1)v → Σperut = n+1 2l Σsimpul =n+2 Σsimpul = Σperut + 1 n = 0,1,2,3……

  5. 2. KolomUdara/Pipaorgana a) Terbuka: a. Frekuensi nada dasar →fo=v/λ→l=1/2λ→λ=2l (Harmonik ke-1) fo= v/2l b. Frekuensi nada atas pertama→f1→v/λ→l=λ (Harmonik ke-2) f1= v/l C. Frekuensi nada atas kedua→f2→v/λ→l=(3/2)λ (Harmonik ke-3) f2 = 3v 2l b) Tertutup : a. Frekuensi nada dasar →fo=v/λ→l=1/4λ→λ=4l (Harmonik ke-1) fo= v/4l b. Frekuensi nada atas pertama→f1→v/λ→l=(3/4)λ (Harmonik ke-2) f1= 3v/4l C. Frekuensi nada atas kedua→f2→v/λ→l=(5/4)λ (Harmonik ke-3) f2 = 5v/4l d. fn=(n+1)v → Σperut = n+1 2l Σsimpul =n+2 Σsimpul = Σperut + 1 n = 0,1,2,3……….. d. fn=(2n+1)v → Σperut = Σsimpul = n+1 4l

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