Question

A 0.30 kg block oscillates back and forth along a straight line on a frictionless horizontal surface. Its displacement from the origin is given by x = (18 cm)cos[(11 rad/s)t + π/2 rad] (a) What is the oscillation frequency? (b) What is the maximum speed acquired by the block? (c) At what value of x does this occur? (d) What is the magnitude of the maximum acceleration of the block? (e) At what positive value of x does this occur? (f) What force, applied to the block by the spring, results in the given oscillation?

Answer #1

Given, x = (18 cm)cos[(11 rad/s)t + π/2 rad]

a.)

Standard equation is given by,

x = A*cos(w*t + π/2)

By comparing it with standard equation,

A = amplitude = 18 cm = 0.18 m

w = 11 rad/s

then, oscillation frequency = f = w/(2*pi)

f = 11/(2*pi)

**f = 1.75 Hz**

b.)

Since,

Maximum speed in oscillation is given by,

Vmax = A*w = 0.18*11

**Vmax = 1.98 m/s**

c.)

V = dx/dt = -A*w*sin(w*t + π/2)

V = -(18 cm)*(11 rad/s)*sin(11*t + π/2)

For maximum value,

sin(11*t + π/2) = 1

**t = 0**

d.)

Acceleration(a) = dV/dt = -A*w^2*cos(w*t + π/2)

So, maximum value will be,

a = A*w^2 = 0.18*11^2

**a = 21.78 m/s^2**

e.)

For maximum value of acceleration,

x = A

**x =
18 cm**

f.)

In oscillation:

w = sqrt(k/m)

k = m*w^2 = 0.30*11^2

k = 36.3 N/m

From hooke's law,

F = -k*x

F = -36.3*x

F = -36.3*(0.18*cos(11t + π/2))

**F = -6.534*cos(11t + π/2))** N

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