A block of mass m is at rest on a smooth horizontal surface under the application of a constant force Fo. A small particle of mass m moving with velocity uo hits the block and gets embedded in it. What will be the frequency & amplitude of oscillation of the block
? Uo k m m →F. 1 1 k m т
(A ) иол 2πV 2m k k 2m (B) 2TV 2m lov 2k m т (C) ио Monk m -16 -15 m 1 (D) k 21 V2m Uo . 2k
Answers
Answer:
Explanation:
Equations
Equation Symbol breakdown Meaning in words
T_s = 2\pi\sqrt{\dfrac{m}{k}}T
s
=2π
k
m
T, start subscript, s, end subscript, equals, 2, pi, square root of, start fraction, m, divided by, k, end fraction, end square root T_sT
s
T, start subscript, s, end subscript is the period of the spring, mmm is the mass, and kkk is the spring constant. The period of a spring-mass system is proportional to the square root of the mass and inversely proportional to the square root of the spring constant.
How to analyze vertical and horizontal spring-mass systems
Both vertical and horizontal spring-mass systems without friction oscillate identically around an equilibrium position if their masses and springs are the same.
For vertical springs however, we need to remember that gravity stretches or compresses the spring beyond its natural length to the equilibrium position. After we find the displaced position, we can set that as y=0y=0y, equals, 0 and treat the vertical spring just as we would a horizontal spring. Figure 1 below shows the resting position of a vertical spring and the equilibrium position of the spring-mass system after it has stretched a distance ddd.
Figure 1. To the left of this image is the resting position of the spring and to the right is the displaced equilibrium position of the spring when the mass is attached. A vertical spring mass system oscillates around this equilibrium position of y=0y=0y, equals, 0.
We can use a free body diagram to analyze the vertical motion of a spring mass system. We would represent the forces on the block in figure 1 as follows:
Figure 2. The forces on the spring-mass system in figure 1.
Then, we can use Newton's second law to write an equation for the net force on the block:
\begin{aligned}\Sigma F &= ma \\ \\ &=F_s - F_g \\ \\ &= kd - mg \end{aligned}
ΣF
=ma
=F
s
−F
g
=kd−mg
The block in figure 1 is not accelerating, so our equation simplifies to:
kd - mg = 0kd−mg=0
Answer:
mk
F
Maximum speed is at mean position(equilibrium). F=kx
x=
k
F
W
F
+W
sp
=ΔKE
F(x)−
2
1
kx
2
=
2
1
mv
2
−0
F(
k
F
)−
2
1
k(
k
F
)
2
=
2
1
mv
2
⇒V
max
=
mk
F