Respuesta :
Answer: T2 = 7.07s
Explanation: The period of a loaded spring of spring constant k and mass m is given by
T= 2π √m/k
With 2π constant and k, it can be seen with little algebra that
T² is proportional to mass m
Hence (T1)²/m1 = (T2) ²/m2
Where T1 = 5, T2 =?, let m1 = m hence m2 = 2m.
By substituting, we have that
5²/m = (T2) ²/2m
25 / m = (T2) ²/2m
25 × 2m = (T2) ² × m
25 × 2 = (T2) ²
50 = (T2) ²
T2 = √50
T2 = 7.07s
The period of the system is T2 = 7.07s
Calculation of the period:
Here The period of a loaded spring of spring constant k and mass m should be provided by
T= 2π √m/k
Now
T² is proportional to mass m
Therefore, (T1)²/m1 = (T2) ²/m2
Now
5²/m = (T2) ²/2m
25 / m = (T2) ²/2m
25 × 2m = (T2) ² × m
25 × 2 = (T2) ²
50 = (T2) ²
T2 = √50
T2 = 7.07s
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