Identical weights suspended from a spring vibrate vertically. When 5 weights were removed, the oscillation period

Identical weights suspended from a spring vibrate vertically. When 5 weights were removed, the oscillation period decreased from 4.5 to 3 seconds. Determine the initial number of weights.

T1 = 4.5 s.

T2 = 3 s.

n -?

The period of free oscillations of a spring pendulum T is determined by the formula: T = 2 * P * √m / √k, where P is the number pi, m is the mass of the load, k is the stiffness of the spring.

T1 = 2 * P * √ (n * m1) / √k, where n is the number of cargoes, m1 is the mass of one cargo.

T2 = 2 * P * √ ((n – 5) * m1) / √k.

Let us express the ratio of the oscillation periods: T1 / T2 = 2 * P * √ (n * m1) * √k / √k * 2 * P * √ ((n – 5) * m1) = √ (n * m1) / √ ( (n – 5) * m1) = √n / √ ((n – 5).

n / (n – 5) = T1 ^ 2 / T2 ^ 2.

n * T2 ^ 2 = (n – 5) * T1 ^ 2.

n * T2 ^ 2 = n * T1 ^ 2- 5 * T1 ^ 2.

n = 5 * T1 ^ 2 / (T1 ^ 2 – T2 ^ 2).

n = 5 * (4.5 s) ^ 2 / ((4.5 s) ^ 2 – (3 s) ^ 2) = 9.

Answer: initially there were n = 9 weights on the spring.