What mass on a spring with a spring constant of 120 N/m will oscillate with a period of 1.0 s?
F = -k x = m a
if x = a sin 2 pi f t
v = 2 pi f a cos 2 pi f t
a = - (2 pi f)^2 a sin 2 pi f t
so
a= -(2 pi f)^2 x
then
-k x = -m (2 pi f)^2 x
(2 pi f)^2 = k/m
here T = 1/f = 1 so f = 1/1 = 1 Hz
(2 pi)^2 = 120/m
m = 30/( pi^2) = 3.04 kg
To solve this question, you can use the formula for the period (T) of an oscillating mass-spring system:
T = 2π√(m/k)
where T is the period, m is the mass, and k is the spring constant.
In this case, we have the period (T) as 1.0 s and the spring constant (k) as 120 N/m. We need to find the mass (m).
Rearranging the formula, we have:
T = 2π√(m/k)
(1.0) = 2π√(m/120)
To solve for m, we can isolate it by squaring both sides of the equation:
(1.0)² = (2π√(m/120))²
1.0 = 4π²(m/120)
1.0 = 4π²(m/120)
Now, we can solve for m by multiplying both sides of the equation by 120 and dividing by 4π²:
m = (1.0 * 120) / (4π²)
Using a calculator, we can evaluate this expression:
m ≈ 0.957 kg
So, a mass of approximately 0.957 kg on a spring with a spring constant of 120 N/m will oscillate with a period of 1.0 s.