A horizontal spring is fixed at one end, then stretched 3.0 cm by a force of 6.0 N.
Afterwards, a 0.50 kg body is attached, pulled 2.0 cm from its rest position, and
released. Find
(a) spring constant,
(b) angular frequency,
(c) frequency and
(d) period.
k = (Force)/(deflection) in N/m
w = sqrt(k/m) in radians/sec
if mass m is in kg
f = 2 pi w
Period = 1/f
You really need to learn these formulas and not have someone keep reminding you.
To find the spring constant, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring.
Hooke's Law can be written as:
F = -kx
Where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.
In this case, the spring is stretched 3.0 cm (0.03 m) by a force of 6.0 N. So we have:
6.0 N = -k * 0.03 m
To find k, we rearrange the equation:
k = -6.0 N / 0.03 m = -200 N/m
Therefore, the spring constant is -200 N/m.
Now let's move on to finding the angular frequency.
The angular frequency (ω) can be found using the formula:
ω = √(k / m)
Where k is the spring constant and m is the mass attached to the spring.
In this case, the mass attached is 0.50 kg, and we have already found the spring constant as -200 N/m.
ω = √((-200 N/m) / 0.50 kg)
= √(-400 N/(kg*m²))
= √(400 N/(kg*m²)) * i (where i is the imaginary unit)
= 20i rad/s
Therefore, the angular frequency is 20i rad/s.
Moving on to finding the frequency.
The frequency (f) can be found using the formula:
f = ω / (2π)
Where ω is the angular frequency.
In this case, the angular frequency is 20i rad/s.
f = (20i rad/s) / (2π)
= 10i / π Hz
Therefore, the frequency is 10i / π Hz.
Lastly, let's find the period.
The period (T) is the reciprocal of the frequency, so we can calculate it using the formula:
T = 1 / f
In this case, the frequency is 10i / π Hz.
T = 1 / (10i / π Hz)
= π / (10i) s
Therefore, the period is π / (10i) s.