A body of mass 20g is suspended from the end of a spiral spring whose force constant is 0.5N/m. The body is set into a simple harmonic motion with amplitude 0.2m. Calculate the period of the motion?
period = 2PI sqrt (m/k) m is in kg, k in N/m
3.9 sec
To find the period of the motion, we can use the formula:
T = 2π * √(m/k)
where:
T = period of the motion
π ≈ 3.14
m = mass of the body in kg
k = force constant of the spring in N/m
First, convert the mass from grams to kilograms:
m = 20g = 0.02kg
Now, substitute the values into the formula:
T = 2π * √(0.02kg / 0.5N/m)
T = 2π * √(0.04N)
T = 2π * 0.2N
T ≈ 1.26 seconds
Therefore, the period of the motion is approximately 1.26 seconds.
To calculate the period of the simple harmonic motion, we can use the equation:
T = 2π√(m/k)
where:
T = period of the motion
m = mass of the body
k = force constant of the spring
Given:
m = 20g = 0.02kg
k = 0.5 N/m
Let's substitute the values into the equation:
T = 2π√(0.02/0.5)
First, we need to divide the mass by the force constant:
0.02 / 0.5 = 0.04
Now, let's substitute this value into the equation:
T = 2π√0.04
Next, take the square root of 0.04:
√0.04 = 0.2
Finally, substitute this value back into the equation and solve for T:
T = 2π * 0.2
T = 1.26 seconds
Therefore, the period of the motion is approximately 1.26 seconds.