Question

the mechanical energy of a mass-spring system is 0.202 j. if the force constant of the spring of 300 n/m and the mass is 0.64 kg, what is the amplitude of the oscillatory motion in cm?

Answer 1

The amplitude of the oscillatory motion for a mass-spring system with a mechanical energy of 0.202 J, a spring constant of 300 N/m, and a mass of 0.64 kg is 3.67 cm.

Step-by-step explanation:

The mechanical energy of a mass-spring system is 0.202 J, and we are tasked with finding the amplitude of the oscillatory motion. The formula for the total mechanical energy in a spring-mass oscillator is given by E = 1/2 k A^2, where E is the total energy, k is the force constant of the spring, and A is the amplitude. Plugging in the given values:

E = 0.202 J

k = 300 N/m

We rearrange the formula to solve for A:

A = sqrt((2 * E) / k) = sqrt((2 * 0.202 J) / 300 N/m) = sqrt(0.404 J / 300 N/m) = sqrt(0.00134667 m^2) = 0.0367 m

Converting to centimeters:

A = 0.0367 m * 100 cm/m = 3.67 cm

Thus, the amplitude of the oscillatory motion is 3.67 cm.