Question

Определите частоту колебаний и коэффициент жесткости пружины, на котором груз массой 200г совершает 60 колебаний за 2мин

Answer 1

Частота колебании ню
ню=60/120=0,5Гц. Т=120/60=2с.
к=4·3,14^2·0,2кг/2^2=1,97 Н/м

Answer 2

Calculation of Frequency and Spring Constant

To determine the frequency of oscillations and the spring constant of a spring on which a mass of 200g performs 60 oscillations in 2 minutes, we can use the following formulas:

1. Frequency (f) is given by the formula: f = (1 / T), where T is the period of oscillation.

2. Period (T) is given by the formula: T = (time / number of oscillations).

3. The spring constant (k) is given by the formula: k = (4π²m) / (T²), where m is the mass of the object attached to the spring.

Let's calculate the frequency and spring constant step by step.

Step 1: Calculate the period (T): - The time given is 2 minutes, which is equal to 120 seconds. - The number of oscillations is 60. - Therefore, the period (T) is: T = (120 seconds / 60 oscillations) = 2 seconds/oscillation. Step 2: Calculate the frequency (f): - The frequency (f) is given by the formula: f = (1 / T). - Substituting the value of T, we get: f = (1 / 2 seconds/oscillation) = 0.5 oscillations/second. Step 3: Calculate the spring constant (k): - The mass of the object attached to the spring is 200g, which is equal to 0.2kg. - The period (T) is 2 seconds. - The spring constant (k) is given by the formula: k = (4π²m) / (T²). - Substituting the values of m and T, we get: k = (4π² * 0.2kg) / (2 seconds)². - Simplifying the equation, we find: k = (4 * 3.14² * 0.2kg) / 4 seconds². - Calculating further, we get: k = 3.14² * 0.2kg / 1 seconds². - The spring constant (k) is approximately equal to: k ≈ 0.985 N/m.

Conclusion

Therefore, the frequency of oscillations for the given system is approximately 0.5 oscillations/second, and the spring constant of the spring is approximately 0.985 N/m.