A block of mass M = 6.20 kg, at rest on a horizontal frictionless table, is atta

Question

A block of mass M = 6.20 kg, at rest on a horizontal frictionless table, is attached to a rigid support by a spring of constant An oscillator consists of a block attached to a spring (k = 370 N/m). At some time t, the position (measured from the systems equilibrium location), velocity, and acceleration of the block are x = 0.0893 m, v = -10.4 m/s, and a = -137 m/s^2. Calculate (a) the frequency of oscillation, (b) the mass of the block, and (c) the amplitude of the motion. (a) Number Units (b) Number Units . (e) Number Units

Explanation / Answer

Given,

K = 370 N/m

x = 0.0893 m

v = -10.4 m/s

a = -137 m/s^2

F = ma

F = -kx

ma = -kx

m = -kx/a (where k, x, and a where previously stated)

m = -370 * 0.0893 / (-137)

mass = 0.241 kg

T = 2 * (pi) * sqrt(m / k)

T = 2 * pi * sqrt(0.241 / 370)

T = 0.1603

f = 1/T

f = 1/0.1603

f = 6.236 Hz

for some t:
x(t) = Acos(wt) = 0.0893
v(t) = -Awsin(wt) = -10.4
you now have two equations with two unknown and are now able to solve for both

A = 0.0893/cos(wt)

A = 10.4/(w*sin(wt))

0.0893/cos(wt) = 10.4/w(sin(wt)

T = 2(pi)/w

0.1603 = 2(pi)/w

w = 39.196

sin(39.196 t)/cos(39.196 t) = 10.4/(39.196 *0.0893)

tan(39.196 t) = 2.97

39.196 t = 1.246 sec

then put this into displacement equation

x(t) = Acos(39.196 t) = 0.0893

A = 0.0893/cos(39.196 t)

A= 0.0893/cos(1.246)

A = 0.279 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.