PR KELOMPOK 1.2. AlIhbsanertBarran us Chandra Wijayanto 1106067886/Tekni 1106067886/T ekni k Mesi n 1106012804/Tekni k Perka al a n 3.4. MMuhammad Hanif NadhiJihf bidz Sininina 1106052524/Tekni k Mesi n 1106054555/Tekni 1106054555/T ekni k Perka al a n 5.6. PetRivraladEloaGarchi ng Pradana 1106023096/Tekni k Perka al a n a 1106000016/Tekni k Mesi n 7. Ruhama Sidqy 1106068806/Teknik Perka alan PR FISDAS 9 BAB 16 : 6, 19, 24, 27, 30, 32, 37 , 42, 50, 55, dan 60
A sinusoidal wave travels long a string under tension. Figure 16-30 gi es the slopes along the string at time t = O. The scale of the x axis is set by Xs = 0.80 m. W at is the amplitude of the wave? 6.
ൌ sinሺ െ ሻ ൌ 2 ൌ 2 cosሺ െ ሻ ൌ ൌ 0.0.42ൌ 5 ିଵ ൌ cosሺ cosሺ െ ሻ ൌ 5 cosሺ5 ሺ0ሻ െ ሺ0ሻሻ ൌ0.013 ൌ 1.3 What is the speed of a tra sverse wave in a rope of length l ength 2.00 m and tension of 500 N? 19.
ass 60.0 g under a
ൌ ඨ / ൌ ඨ 0.50006/2 ൌ129 / In Fig. 16-35a, string 1 h s a linear density of 3.00 g/m, and string 2 as a linear density of 5.00 g/m. They are under ension due to the hanging block of mass M = 500 g. Calculate the wave speed speed on (a) string 1 and (b) string 2. (Hint: When a string loops halfway around a pul-ley, it pulls on the pulley with a net force that is twice the tension in t he string.) Next the block is divided into two two blocks (with Ml + Mz = M) and the apparat us is rearranged as shown in Fig. 16-35b. Find ( ) Mj and (d) Mz such that the wave wave speeds in the two strings are equal. a) The tension in each st ring is given by τ = Mg /2. Thus, the wave wave peed in string 1 is 24.
b) And the wave speed i string 2 is
c) Let We solve for M 1 and obtain
d) And we solve for the s econd mass: M 2 = M – M 1 = (500 g – 187.5 g ) ≈ 313 g.
27.
We note from the graph ( nd from the fact that we are dealing with a c sine squared, see
Eq. 16-30) that the wave freq ency is f =
ଵ ଶ௦
= 500 Hz, and that the wavel ngth λ = 0.20 m.
We also note from the graph t hat the maximum value of dK/dt is 10 W. Se ting this equal to the maximum value of Eq. 16 -29 (where we just set that cosine term equal to 1) we find
with SI units understood. ubstituting in μ = 0.002 kg/m, ω = 2π f and v = f λ , we solve for the wave amplit de:
Use the wave equation to find the speed of a wave given in terms of t he general function hex, t): -1 -1 y(x, t) = (4.00 mm) h[(30 m ) x + (6.0 S )t]. 30.
ൌ ൌ
60 30
ൌ 0.2
/
What phase difference direction along a stretched s times that of the common am degrees, (b) radians, and (c) 32.
a) ∅
ൌ2
ିଵ ଵ.ହ௬ ሺ ሻ ଶ௬
ൌ 82,9˚ b)∅ ൌ 82,9˚/57˚ ൌ 1,45
etween two identical traveling waves, m ving in the same ring, results in the combined wave having an amplitude 1.50 litude of the two combining waves? Expres your answer in (a) avelengths.
c)
ʎൌ5,ଵ,72ସ25ହ ൌ
37.
These two waves travel along the same string:
Y1 (x, t) = (4.60 mm) sin(2 x - 400 t) Y2(X, t) = (5.60 mm) sin(2 x - 400 t + 0.80 rad). What are (a) the amplitude and (b) the phase angle (relative to wave 1) of the resultant wave? (c) If a third wave of amplitude 5.00 mm is also to be sent along the string in the same direction as the first two waves, what should be its phase angle in order to maxi-mize the amplitude of the new resultant wave? 0
1
= 4,60 mm;
=0
2
= 5,60 mm;
= 0.80 = 144
0
Fm2
a. Amplitude: Penambahan vektor 0
Y’mh = ym1 cos0 + ym2 cos 144
0.80
0
Fm1
= 4,6 (1) + 5,6 (-0,809) = 0,6950 mm 0
Y’mv = ym1 sin 0 + ym2 sin 144
0
= 4,6 (0) + 5,6 (0,5877) = 3,2916 mm Y’
=
ඥ ሺ0,6950ሻଶ +ሺ3,2916ሻଶ
=3,29mm
b.
=
arctan ଷ,,ଶଽଵଽହ
= 1,36 rad c. Fase yang terbentuk adalah sama untuk berapapun resultan karena sudut fase awal tidak berubah A string under tension i oscillates in the third harmonic at fre-quency f 3, and the waves on the string have wavelength 3 If the ten-sion is increased to f = 4 i and the string is again made to oscillate in the third harmonic, what then are (a) the frequency of oscillation in terms of f 3 and (b) the wavelength of the waves in terms of 3? 42.
Frekuensi resonansi:
v=
ඥ /
; f n =
௩ଶ ଶ ට ఓఛ =
a. Ketika f’3 =
f =
ଷଶ ට ఛఓ
4
i
= 2f 3
b. Panjang gelombang baru 3=
′
50.
௩ ଶଷ య ′ ′
=
=
3
Dik : x =0, y(x,y) = -(0,04) cos kx sin wt W=2
/
= π rad/s
Dit : a. y (t =0,50 s) = .....? b. y (0,30 m , 0,50 s) =......? c. u ( 0,50 s , 0,20 m ) =.....? d. u (t =1s) =........? e. daerah fungsi saat t =0,50s ? Jawab : a. y (0,20m , 0,50s) = -(0,04) cos kx sin wt = 0,040 m b. y (0,30m , 0,50s) = -(0,04) cos kx sin wt = 0 m c. u =
ௗ௬ௗ௫
=-(0,04) cos kx sin wt = 0 m
d. u =-0,13 m saat t =1s e. daerah fungsi saat t =0,50s adalah
0≤ ≤40
Diketahui amplitudenya: 12cm. ω dan frekuensi dapat ditemukan dengan membandingkan dengan persamaan dasar y = ym sin (k x ± ω t ). Anti-node bergerak sepanjang 12 cm dalam suatu gerakan harmonis sederhana. Kemudian, karena periode T berhubungan dengan frekuensi angular, maka : 55.
T =2π / ω = 2π /4.00π =0.500s
Maka, pada saat t =½T =0.250s, gelombangnya bergerak sejauh ∆ x=vt dimana kecepatan gelombangnya v = ω / k = 1.00 m/s. Maka, ∆ x= (1.00 m/s) (0.250 s) = 0.250 m.
In Fig. 16-41, a string, tied to a sinusoidal oscillator at P and running over a support at Q, is stretched by a block of mass m. The separation L between P and Q is 1.20 m, and the frequency f of the oscillator is fixed at 120 Hz. The amplitude of the motion at P is small enough for that point to be considered a node. A node also exists at Q. A standing wave appears when the mass of the hanging block is 286.1 g or 447.0 g, but not for any intermediate mass. What is the linear density of the string? 60.
-> Dengan string tetap pada kedua ujungnya, dengan menggunakan Persamaan. 16-66 dan Persamaan. 16-26, frekuensi resonansi dapat ditulis sebagai
ൌ 2 ൌ 2 ඨ ൌ 2 ඨ µ
ൌ1,2,3,…
g
µ
Massa yang memungkinkan osilasi untuk mengatur harmonik nth pada string adalah
4ൌ ଶଶ ଶ
µ
g
Dengan demikian, kita melihat bahwa massa blok berbanding terbalik dengan jumlah harmonik kuadrat. Dengan demikian, jika blok gram 447 sesuai dengan jumlah harmonik n, maka
karena itu,
ଶ ଶ ሺ ሻ 447 +1 + 2 + 1 2 +1 ൌ ൌ ൌ1+ ଶ ଶ ଶ 286. 1 ସ଼ 5 624 ሺ2 +1ሻ ଶ଼.ଵሺെ1ൌ0. ଶሻ harus sama bilangan ganjil
dibagi dengan kuadrat
bilangan bulat . yaitu, mengalikan 0,5624 oleh persegi (seperti 1, 4, 9, 16, dll) harus memberi kita jumlah yang sangat dekat (dalam ketidakpastian eksperimental) harus angka ganjil (1, 3, 5, ...). Mencoba ini dalam suksesi (dimulai dengan perkalian dengan 1, kemudian oleh 4, ...), kita menemukan bahwa perkalian dengan 16 memberikan nilai yang sangat dekat dengan 9, kami menyimpulkan n = 4, (sehingga n2 = 16 dan 2n + 1 = 9). Mencolokkan m = 0,447 kg, n = 4, dan nilai-nilai lain yang diberikan dalam masalah, kita menemukan µ
ൌ0.000845 ⁄ ൌ0.845 ⁄ kg m
g m
