The 0.18 kg block oscillates back and forth along a straight line on a frictionless horizontal surface. Its displacement from the origin is given by the following equation. x = (11 cm) cos[ (15 rad/s) t + π/2 rad) ] (a) What is the oscillation frequency? 2.39 Correct: Your answer is correct. Hz (b) What is the maximum speed acquired by the block? 165 Correct: Your answer is correct. cm/s At what value of x does this occur? cm (c) What is the maximum acceleration of the block? cm/s2 At what values of x does this occur? (positive then negative) cm and cm (d) What force, applied to the block, results in the given oscillation? (N/m) x

Question

Physics

Kaylee Swanson

25 September, 23:54

The 0.18 kg block oscillates back and forth along a straight line on a frictionless horizontal surface. Its displacement from the origin is given by the following equation. x = (11 cm) cos[ (15 rad/s) t + π/2 rad) ] (a) What is the oscillation frequency? 2.39 Correct: Your answer is correct. Hz (b) What is the maximum speed acquired by the block? 165 Correct: Your answer is correct. cm/s At what value of x does this occur? cm (c) What is the maximum acceleration of the block? cm/s2 At what values of x does this occur? (positive then negative) cm and cm (d) What force, applied to the block, results in the given oscillation? (N/m) x

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Answers (1)

Know the Answer?

Short · Answer 1

(a) f = 2.39 Hz

(b) Vmax = 165 cm/s

(c) amax = 24.75 m/s²

Explanation:

Given:

Mass, m = 0.18kg

displacement from the origin = (11 cm) cos [ (15 rad/s) t + (π / 2 rad) - 1

(a)

15 rad/s will be the angular frequency, ω

So frequency is

f = ω / 2π

f = 15 / 2π

f = 2.39 Hz

(b)

we know,

Vmax = ω xmax

From equation 1, the value of xmax is 11cm

xmax = 0.11m

Vmax = 15 rad/s X 0.11m

Vmax = 1.65 m/s

Vmax = 165 cm/s

(c)

Maximum acceleration, amax = ω²xm

amax = (15) ² X 0.11m

amax = 24.75 m/s²