1. The masaafada Inta u dhaxaysa labada god ee hirarka dusha sare ee biyuhu waa 20 m. Shay ayaa dul sabeynaya biyaha si uu u dareemo dhaqdhaqaaqa gariirka. Haddii waqtiga lagu socdo hal gariir uu yahay 4 ilbiriqsi, markaa xawaaraha hirarku waa .... m/s
A. 20
B. 15
C. 10
D. 5
La yaqaan:
Milanello (λ) = 20 meters
Period (T) = 4 seconds
SE buska: Xawaaraha hirarka (v)
Xalka:
Isle'egta xawaaraha hirarka:
v = λ / T = 20 meters / 4 seconds = 5 meters / second
Jawaabta saxda ah waa D.
2. Two points A and B are on the rope and are 90 cm apart from one another. On the rope propagates the transverse waves, so that point A is at the top of the wave, point B at the bottom of the wave, and between them, there are two crests and two troughs. If the wave period is 0.3 seconds, then the wave propagation is…
A. 10.8 cm/s
B. 18.0 cm/s
C. 120.0 cm/s
D. 200.0 cm/s
La yaqaan:
Distance AB (l) = 90 cm
Period of wave (T) = 0.3 seconds
Frequency of wave (f) = 1/0.3 seconds
Point A is at the crest of the wave and point B is at the trough of the wave.
Between point A and B, there are two crests and two troughs.
La doonayo: Speed of wave (v = f λ)
Xalka:
Based on figure, can conclude there are 2.5 wavelengths. Distance of a wavelength (λ) = 90 cm / 2.5 = 36 cm
Xawaaraha hirarka:
v = f λ = (1 / 0.3)(36) = 36 / 0.3 = 120 cm/s.
Jawaabta saxda ah waa C.
3. The following graph displays the displacement of a point in one medium as a function of time when a wave passes through the medium.
If the wavelength is 6 meters, then the speed of the wave propagation is…
A. 3 m/s
B. 6 m/s
Q. 8 m/s
D. 12 m/s
La yaqaan:
There are two wavelengths based on the graph above.
Distance of 1 wavelength (λ) = 6 meters / 2 = 3 meters
Period (T) = 0.5 seconds
La doonayo: Speed of waves (v)
Xalka:
v = f λ = λ / T = 3 meters / 0.5 seconds = 6 meters / second
Jawaabta saxda ah waa B.
4. Based on the figure below, the point that has phase difference ¾ with point A is ….
A. Point B
B. Point C
C. Point D
D. Point E
Solution
Point B has a phase difference of ¼ λ with point A
Point C has a phase difference of 2/4 λ or 1/2 λ with point A
Point D has a phase difference of 3/4 λ with point A
Point F has a phase difference of 5/4 λ with point A.
Jawaabta saxda ah waa C.
5. On a pond’s surface, there are two dry leaves 60 centimeters away from each other. Both move up and down like the surface of the water with a frequency of 2 Hz. When one leaf is at the crest, the other leaves are at the trough, and between them, there is one crest and one trough. Determine the speed of the wave propagation of the wave.
A. 20 cm/s
B. 30 cm/s
C. 80 cm/s
D. 120 cm/s
La yaqaan:
Distance between both leaves = 60 cm
Frequency (f) = 2 Hz = 2
La doonayo: The speed of wave
Xalka:
Between both leaves, there are 1.5 wavelengths. Distance of 1 wavelength is (λ) = 60 cm / 1.5 = 40 cm
Speed of wave (v) :
v = f λ = (2 Hz)(40 cm) = 80 cm/second
Jawaabta saxda ah waa C.
6. Based on the figure below, determine the amplitude, period, frequency, and speed of the wave.
Xalka:
Amplitude (A) = 4 meters
Period (T) = 6 seconds / 3 = 2 seconds
Frequency (f) = 1 / T = 1 / 2 = 0.5 hertz
Wavelength (λ) = 24 meters / 3 = 8 meters
Speed of wave (v) = f λ = (0.5 hertz)(8 meters) = 4 meters/second or
Speed of wave (v) = λ / T = 8 meters / 2 second = 4 meters/second
Jawaabta saxda ah waa D.
7. On a string with length of 1.2 m and mass of 200 g formed 1.5 sinusoidal waves with frequency of 50 Hz. Based on these data, determine the wave period and the tension force of the rope.
A. Period = 0.02 seconds and tension force = 6.67 N
B. Period = 0.01 seconds and tension force = 6.67 N
C. Period = 0.02 seconds and tension force = 266.67 N
D. Period = 0.01 seconds and tension force = 266.67 N
La yaqaan:
Length of rope (l) = 1.2 meters and there are 1.5 sinusoidal waves so that distance of 1 wavelength (λ) = 1.2 meters / 1.5 = 0.8 meters
Mass of rope (m) = 200 gram = 0.2 kg
Soo noqnoqoshada (f) = 50 Hz
Density of rope (µ) = m/l = 0.2 kg / 1.2 meters = (1/6) kg/meter
La doonayo: Period of wave (T) and the tension force of rope (T)
Xalka:
Period of wave :
T = 1 / f = 1 / 50 Hz = 0.02 seconds
The speed of wave on rope :
v = f λ = (50 Hz)(0.8 meters) = 40 meters/second
Xoogga xiisadda (T):
Jawaabta saxda ah waa C.
