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.