Question

A spring with spring constant 14.9 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is then pulled down 5.00 cm and released. The ball makes 36.0 oscillations in 19.0 seconds.

Part A
What is the mass of the ball? in g.

Part B
What is the maximum speed?in cm/s.

Answer 1

let theta be @
k = 14.9 N/m , A = 0.05 m , @ = 3(2p) = 72p , t = 19 s

part A

w = @/t = 11.9
k = mw²
14.9 = m x 11.9²
m = 0.1052 Kg = 105.2 g

part B
Maximum speed, v = Aw = 0.595 m/s = 59.5 cm/s

Answer 2

A spring with spring constant 14.9 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is then pulled down 5.00 cm and released. The ball makes 36.0 oscillations in 19.0 seconds.

Hooke's Law = F = -kx
x is the displacement of the spring's end from its equilibrium position (a distance, in SI units: metres);
F is the restoring force exerted by the spring on that end (in SI units: N or kg·m·s-2); and
k is a constant called the rate or spring constant (in SI units: N·m-1 or kg·s-2).

So F = -14.9 x -0.05 m = 0.745 N-m (newton-meters)

Note: "-0.05" because it's pulled down (aka negative) and it's in centimeters that we convert to meters (aka divide by 100)

Maximum speed? Vmax = A*ω where A = .05 m and ω = √(N/m) = 3.86 N/M/sec
Ergo, Vmax = 0.05 x 3.86 = 0.193 M/sec

Answer 3

Use Hooke's Law to find the mass:

F = -kx
mg = -kx
m = -kx/g
m = kx/g (we can ignore the - due to sign convention and we know mass must be +)
m = (14.9N/m)(0.05m)/(9.8m/s²)

m = 0.0760 kg = 76 g

To find the max speed, simply use the equation v_max = ωA, where ω is angular velocity and A is amplitude.

We know:

ω = 2πf or you could know that ω = √(k/m), we'll perform work on the former:

Frequency is simply cycles/second, in this case 36/19, therefore ω = 2π(36/19) and A = 0.05 m

Therefore,

v_max = ωA = 2π(36/19)(0.05)

v_max = 0.595 m/s or 59.5 cm/s