
При игре на гитаре середина струны проходит расстояние 0,5 см между крайними положениямиха 0,001 с.
Определите амплитуду , частоту и период колебаний струны.

Ответы на вопрос

Период за какое время возвращается в первоначальное положение =0,001 с*2=0,002 с, так как это половина периода,
Частота= 1/период= 500 Гц



Amplitude, Frequency, and Period of String Oscillations
To determine the amplitude, frequency, and period of string oscillations, we need to understand the properties of the string and the motion it undergoes.
Amplitude refers to the maximum displacement of the string from its equilibrium position. In this case, the string oscillates between two extreme positions, with a distance of 0.5 cm between them. Therefore, the amplitude of the string oscillations is 0.5 cm.
Frequency is the number of complete oscillations or cycles that occur in one second. It is measured in hertz (Hz). To calculate the frequency, we need to know the time it takes for one complete oscillation.
Given that the time taken for the string to move between the extreme positions is 0.001 s, we can calculate the frequency using the formula:
Frequency (f) = 1 / Time period (T)
The time period is the time taken for one complete oscillation. Therefore, the time period (T) can be calculated as:
T = 2 * Time taken for half oscillation
In this case, the time taken for half oscillation is 0.001 s. Therefore, the time period (T) is:
T = 2 * 0.001 s = 0.002 s
Now, we can calculate the frequency using the formula mentioned earlier:
f = 1 / T = 1 / 0.002 s = 500 Hz
Hence, the frequency of the string oscillations is 500 Hz.
Period is the time taken for one complete oscillation. It is the reciprocal of frequency. Therefore, the period (T) can be calculated as:
T = 1 / f = 1 / 500 Hz = 0.002 s
Hence, the period of the string oscillations is 0.002 s.
To summarize: - Amplitude: 0.5 cm - Frequency: 500 Hz - Period: 0.002 s
Please note that these calculations are based on the given information and assumptions about the string's behavior.


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