1. Ntho e thothomela ka maqhubu a 5 Hz ho ya ka ho le letona le ka ho le letshehadi. Ntho e tsamaya ho tloha ntlheng ya tekano ho ya ho maximum ho baleha ka ho le letona. Fumana nako e hlokahalang ho fihlella sebaka se phahameng ka ho fetisisa sa ho falla ka ho le letona ka makhetlo a leshome le motso o mong.
A. 2.05
B. 2.20
C. 2.25
D. 2.50
Tse tsejoang:
Frequency (f) = the amount of vibration for 1 second = 5 Hz
Period (T) = the time interval to do one vibration = 1/f = 1/5 = 0.2 seconds
Oa Batla: The time interval required to reach the maximum displacement at rightward eleven times
Tharollo:
The pattern of the object vibration :
(1 vibration) : B → C → B → A → B
For one vibration, the object performs four vibrations that are B to C, C to B, B to A, A to B. The time interval required for a single vibration is 0.2 seconds / 4 = 0.05 seconds.
The time interval required to reach the maximum displacement at rightward eleven times = (10 x 0.2 seconds) + 0.05 seconds = 2 seconds + 0.05 seconds = 2.05 seconds.
Karabo e nepahetseng ke A.
2. A spring is hung with an object and vibrated. For the vibration frequency to double the original vibration frequency, then the mass of the object is changed to…
A. twice the mass of the original load
B. four times the mass of the original load
C. half the load mass time
D. a quarter of the original load mass
Tharollo:
The equation of the frequency of the spring’s vibration :
f = frequency, k = constant, m = mass of object
Frequency (f) of spring’s vibration if k = 1 time, m = 1 time :
For the frequency of (f) the spring vibration to be twice the mass (m) is changed to 0.25 times or 1/4 times:
Karabo e nepahetseng ke D.
3. A spring with a constant k = 1000 N / m is hung with an object with a mass of 400 grams. The object is pulled to the right as far as 5 cm, then released, so the object is simple oscillating harmonics. Determine the amplitude and frequency of the object oscillation.
Tse tsejoang:
Spring’s constant (k) = 1000 N/m
Boholo of object (m) = 400 gram = 0.4 kg
Amplitude (Δx) = 5 cm = 0.05 m
Ho batloa: Amplitude and frequency of oscillation
Tharollo:
The amplitude of spring oscillation = Δx = 5 cm.
The frequency of oscillation :
Karabo e nepahetseng ke B.
