Mwendo rahisi wa harmonic - matatizo na suluhisho

1. Kitu hutetemeka kwa masafa ya 5 Hz kuelekea kulia na kushoto. Kitu husogea kutoka sehemu ya usawa hadi kiwango cha juu zaidi makazi yao upande wa kulia. Amua muda unaohitajika ili kufikia kiwango cha juu cha uhamishaji kuelekea kulia mara kumi na moja.

A. 2.05

B. 2.20

C. 2.25

D 2.50

Inajulikana:

Frequency (f) = the amount of vibration for 1 second = 5Hz

Period (T) = the time interval to do one vibration = 1/f = 1/5 = 0.2 seconds

SE basi: The time interval required to reach the maximum displacement at rightward eleven times

Suluhisho:

Simple harmonic motion - problems and solutions 1The 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.

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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.

Jibu sahihi ni 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

Suluhisho:

The equation of the frequency of the spring’s vibration :

Simple harmonic motion - problems and solutions 2

f = frequency, k = constant, m = mass of object

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Frequency (f) of spring’s vibration if k = 1 time, m = 1 time :

Simple harmonic motion - problems and solutions 3

For the frequency of (f) the spring vibration to be twice the mass (m) is changed to 0.25 times or 1/4 times:

Simple harmonic motion - problems and solutions 4

Jibu sahihi ni 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.

Simple harmonic motion - problems and solutions 5

Inajulikana:

Spring’s constant (k) = 1000 N/m

Misa of object (m) = 400 gram = 0.4 kg

Amplitude (Δx) = 5 cm = 0.05 m

Inayohitajika: Amplitude and frequency of oscillation

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Suluhisho:

The amplitude of spring oscillation = Δx = 5 cm.

The frequency of oscillation :

Simple harmonic motion - problems and solutions 6

Jibu sahihi ni B.

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