Final answer:
To achieve a maximum acceleration of g when the object oscillates with an amplitude of 4.50 cm, the spring constant (k) should be approximately 984.25 N/m.
Step-by-step explanation:
To determine the spring constant (k) that will result in a maximum acceleration of g, we can use the equation for the maximum acceleration in Simple Harmonic Motion (SHM):
a_max = -ω^2A
where a_max is the maximum acceleration, ω is the angular frequency, and A is the amplitude of the oscillation. In this case, we are given that the maximum acceleration should be equal to g, which is approximately 9.8 m/s^2, and the amplitude is given as 4.50 cm, which is equivalent to 0.045 m. Substituting these values into the equation, we get:
a_max = -ω^2(0.045)
Since the maximum acceleration should be equal to g, we can set a_max equal to 9.8 m/s^2:
9.8 = -ω^2(0.045)
Rearranging the equation, we get:
ω^2 = -9.8/0.045
ω^2 ≈ -217.78
Since the angular frequency squared cannot be negative in this context, we can use the absolute value of -217.78, which is 217.78. The angular frequency (ω) is given by:
ω = 2πf
where f is the frequency. The frequency is related to the angular frequency by the equation:
f = 1/T
where T is the period of the oscillation. In this case, we can use a period of 1 second, which is a common choice. Substituting this into the equation for ω:
ω = 2π(1)
ω ≈ 6.28
Now, we can determine the spring constant (k) using the equation:
k = mω^2
where m is the mass of the object attached to the spring. In this case, we are given that the mass is 25 kg. Substituting the values into the equation:
k = (25)(6.28^2)
k ≈ 984.25 N/m
Therefore, the spring constant (k) should be approximately 984.25 N/m to achieve a maximum acceleration of g when the object oscillates with an amplitude of 4.50 cm.