1. A mass m at the end of a spring vibrates with a frequency of 0.88 Hz. When an additional 600 g
mass is added to m, the frequency is 0.60 Hz. What is the value of m?
ω=
√
√
k
1 k
→ f=
m
2π m
f2
(0.6 Hz)
m
m
=
→
=
f1
m2
m+0.600 kg
(0.88 Hz)
√
√
→ m=0.52 kg.
2. It takes a force of 80.0 N to compress the spring of a popgun 0.20 m to load a 0.15 kg ball. With
what speed will the ball leave the gun?
k =F / Δ x=(80.0N)/(0.20m)=400 N /m
The ball will leave at the equilibrium position, where the kinetic energy is maximum. The maximum
potential energy equals the maximum kinetic energy, therefore we have:
1
1
KE Max = m v 20 =PE Max = kA 2 ;
2
2
1
1
(0.150 kg) v 20= (400 N / m)(0.20 m)2
2
2
→ v 0=10.3 m/ s.
3. A 0.5 kg mass vibrates according to the equation x=0.45cos(8.4t), where x is in meters and t is
in seconds. Determine
A=0.45 m
a) the amplitude
b) the frequency
2 π f =8.4 s−1 → f =1.34 Hz
c) total energy
v 0=ω A=(8.4 s−1)(0.45m)=3.78 m/ s
1
1
E= KE Max = mv 20 = (0.5 kg)(3.78 m/ s)2=3.6 J.
2
2
d) potential energy when x=0.3 m.
1 2 1
1
2
2
−1 2
2
PE= k x = (m ω ) x = (0.5 kg)(8.4 s ) (0.3 m) =1.6 J.
2
2
2