A 190 g air-track glider is attached to a spring. The glider is pushed in 9.8 cm and released. A student with a stopwatch finds that 14 oscillations take 18.5 sec.
What is the spring constant?
frequency=1/2pi *sqrt(k/mass)
and frequency is 1/period.
frequency= oscillations/second
f=14osc/18.5sec=0.76Hz
T= 1/f
T= 2pi/w
1/0.76Hz=2pi/w
cross multiply
w=2pi*0.76= 4.775 rad/s
w=sqrt(k/m)
square both sides
w^2=k/m
w^2 * m = k
k = w^2 * m
you have "w" and you have mass. Plug them in and find the spring constant. Don't forget to convert the mass from grams to kilograms.
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 from its equilibrium position.
The equation for Hooke's Law is:
F = -kx
Where F is the force exerted by the spring, k is the spring constant, and x is the displacement of the spring.
In this case, the glider is pushed 9.8 cm, which is equal to 0.098 m. The force exerted by the spring can be calculated by multiplying the mass of the glider (0.190 kg) by the acceleration due to gravity (9.8 m/s^2):
F = (0.190 kg)(9.8 m/s^2) = 1.862 N
We also know that the glider undergoes 14 oscillations in 18.5 seconds. The time for one complete oscillation (T) can be calculated by dividing the total time (18.5 seconds) by the number of oscillations (14):
T = 18.5 s / 14 = 1.321 s
Next, we can calculate the frequency (f) of the oscillations by taking the reciprocal of the time:
f = 1 / T = 1 / 1.321 s = 0.756 Hz
The angular frequency (ω) can be calculated by multiplying the frequency by 2π:
ω = 2πf = 2π(0.756 Hz) ≈ 4.747 rad/s
Now, we can calculate the spring constant (k) using the formula:
k = (4π^2m) / (ω^2x)
k = (4π^2 * 0.190 kg) / (4.747 rad/s)^2 * 0.098 m
k = 73.229 N/m
Therefore, the spring constant is approximately 73.229 N/m.