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1)An object of mass m = 0.25 kg has a horizontal spring attached to its left sid

ID: 1268070 • Letter: 1

Question

1)An object of mass m = 0.25kg has a horizontal spring attached to its left side, and

slides along a frictionless surface. The spring constant is ? = 0.4 N . At t = 0s, the m

object is displaced 0.1m to the right of its equilibrium position. Its initial velocity is 0.4 m , toward the right.


a) What is the period T of the motion?

b) What is the angular frequency ? ?

c) What is the frequency ? ?

d) What is the total energy E ?

e) What is the amplitude A of the motion? (careful!)

f) What is the phase angle ? ?

g) What is the maximum velocity vmax? (careful!)

h) What is the maximum acceleration amax ?

i) What are the position, velocity, and acceleration at t = ? s?

2)A simple pendulum has a period of 2.40s at a place where g = 9.810 m/s^2 What is the
value of g at another place where this pendulum has a period of 2.41s?

3)A wave travels along a string of linear mass density

Explanation / Answer

Given Data
mass m = 0.25 kg,
spring constant k = 0.4 N/m

at t=0
x = 0.1 m,
initial velocity = 0.4 m/s

a) T = 2pi *sqrt(m/k) = 2pi*sqrt(0.25 / 0.4)
T=6.28*sqrt(0.625)

T = 4.964 s

b )

w = 2pi/Time peroid=6.28/4.964

w = 1.265 rad/s

c)f = 1 / Time peroid = 1 / 4.964 = 0.201 Hz

d) E = 0.5kx^2 + 0.5mv^2

= 0.5 * 0.4 * 0.12 + 0.5 * 0.25 *0.42

E = 0.022 J

e) 0.5 * k * A2 = E

A = sqrt (2 * E / k) = sqrt (2 * 0.022 / 0.4)

A = 0.3316 m

g) 0.5 * m * Vmax2 = E

Vmax = sqrt( 2 * E / m) = sqrt( 2 * 0.022 / 0.25)

Vmax = 0.42

h) a = k * A / m = 0.528 m/s
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2)
Time peroid T1=2.40 sec
g1=9.810 m/s^2
T2 = 2.41 sec
g2=?
we know time period T=2pi*sqrt(l/g)
T1/T2 = sqrt(g2/g1)
g2 =g1*(T1/T2)^2 = 9.728 m/s^2

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3)

y(x,t) = 0.27cos(4.5x-30t+?/12)

dy/dt(Partial Derivative) = 0.27*(-30)*(-sin(4.5x-30t+?/12))

d2y/dt2 (Partial) = -0.27*900*cos(4.5x-30t+?/12)

Putting x=0.025m, t=0.07s

acceleration = d2y/dt2 = -37.491

Mass of element =