Question

An object oscillates such that its position x as a function of time t obeys the equation x = (0.222 m) sin(314 s-1 t), where t is in seconds. (a) In one period, what total distance does the object move? (b) What is the frequency of the motion? (c) What is the position of the object when t = 1.00 s?

Answer 1

Answer:

Part a)

$d = 0.888 m$

PART B)

$f = 50 Hz$

PART C)

$x = 0$

Explanation:

PART A)

total distance moved in one complete oscillation is equivalent to 4 times its amplitude

so we have

A = 0.222 m

now the distance moved is given as

$d = 4A$

$d = 4(0.222)$

$d = 0.888 m$

PART B)

as we know that angular frequency of the motion is given as

$\omega = 314 rad/s$

now we have

$\omega = 2\pi f$

$2\pi f = 315$

$f = 50 Hz$

PART C)

as we know that position of the object is given as

$x = 0.222 sin314 t$

so we will have at t = 1 s

$x = 0.222 sin(314 \times 1)$

$x = 0$