Pre-Test 04b

1.A body of mass 5.0 kg is suspended by a spring which stretches 10 cm when the mass is attached. It is then displaced downward an additional 5.0 cm and released. Its position as a function of time is approximatelya.y = 0.10 sin 9.9t

b.y = 0.10 cos 9.9t

c.y = 0.10 cos (9.9t + .1)

d.y = 0.10 sin (9.9t + 5)

e.y = 0.05 cos 9.9t

y = Acos(t + )A = -0.05 my = -0.05cos(t + )

When t = 0 A = -0.05-0.05 = -0.05cos() = 0y = -0.05cos(t)

F = -ky-mg = k(0.10)k = (5)(9.8)/0.1 = 490w = sqrt(k/m) = sqrt(490/5) = 9.9

y = -0.05cos(9.9t)

2.A body oscillates with simple harmonic motion along the x axis. Its displacement varies with time according to the equation x = 5.0 cos (t). The magnitude of the acceleration (in m/s2) of the body at t = 1.0 s is approximatelya.3.5

b.49

c.14

d.43

e.4.3

a = -2(5.0)cos(t)a = -2(5.0)cos(t)a = -2(5.0)cos() = 49 m/s2

3.A body oscillates with simple harmonic motion along the x axis. Its displacement varies with time according to the equation x = 5 sin (t + /3). The phase (in rad) of the motion at t = 2 s isa.7/3

b./3

c.

d.5/3

e.2

Phase is t + /3t = 2Phase is + /3 = 7/3

4.If y = 0.02 sin (30x 400t) (SI units), the frequency of the wave isa.30 Hz

b.15/ Hz

c.200/ Hz

d.400 Hz

e.800 Hz

T = 1/f = 2/ = 400f = /2 = 400/2 = 200/

5.A mass m = 2.0 kg is attached to a spring having a force constant k = 290 N/m as in the figure. The mass is displaced from its equilibrium position and released. Its frequency of oscillation (in Hz) is approximately

a.12

b.0.50

c.0.010

d.1.9

e.0.080

T = 1/f = 2/ = sqrt(k/m) = sqrt(290/2) = 12.04f = /2 = 12.04/2 = 1.9

6.The oscillation of the 2.0-kg mass on a spring is described by where x is in centimeters and t is in seconds. What is the force constant of the spring?a.4.0 N/m

b.0.80 N/m

c.16 N/m

d.32 N/m

e.2.0 N/m

= sqrt/k/m) = 4.0k/m = 16k = 2 x 16 = 32 N/m

7.The motion of a particle connected to a spring is described by x = 10 sin (t). At what time (in s) is the potential energy equal to the kinetic energy?a.0

b.0.25

c.0.50

d.0.79

e.1.0

mv2 = kx2 m(-10sin(t))2 = k(10cos(t))2(-10sin(t))2 = (k/m)(10cos(t))2(-10sin(t))2 = 2(10cos(t))2(sin(t))2 = (cos(t))2sin(t) = cos(t)sin() = cos() = 0.707This is true when t = 0.25 s

8.At sea level, at a latitude where , a pendulum that takes 2.00 s for a complete swing back and forth has a length of 0.993 m. What is the value of g in m/s2 at a location where the length of such a pendulum is 0.970 m?a.0.098 3

b.3.05

c.9.57

d.10.0

e.38.3

T = 2sqrt(L/g)2 = 2sqrt(.97/g)g = 9.5 m/s2

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