Question

A mass is attached to the end of a spring and set into oscillation on a horizontal frictionless surface by releasing it from a compressed position. The record of time is started when the oscillating mass first passes through the equilibrium position, and the position of the mass at any time is described by x = (3.9 cm)sin[(9.3 rad/s)πt]. Determine the following.
(a) frequency of the motion Hz.
(b) period of the motion.
(c) amplitude of the motion cm.
(d) first time after t = 0 that the object reaches the position x = 2.6 cm

Answer 1

Answer:

Explanation:

x = (3.9 cm) Sin(9.3 πt)

Compare with the standard equation

x = A Sin(2πft)

By comparing, A = 3.9 cm

2πf = 9.3π

f = 4.65 Hz

(a) frequency, f = 4.65 Hz

(b) Period, T = 1/f = 1/ 4.65 = 0.22 second

(c) Amplitude, A = 3.9 cm

(d) x = 2.6 cm

2.6 = 3.9 Sin(9.3πt)

0.667 = Sin (9.3πt)

0.73 = 9.3 x 3.14 x t

t = 0.025 second

Answer 2

Answer

given,

x = (3.9 cm)sin[(9.3 rad/s)πt]

general equation of displacement

x = A sin ω t

A is amplitude

now on comparing

c) Amplitude  =3.9 cm

a) frequency =

     $f = \dfrac{\omega}{2\pi}$

     $f = \dfrac{9.3\pi}{2\pi}$

           f = 4.65 Hz

b) period of motion

        $T= \dfrac{1}{f}$

        $T= \dfrac{1}{4.65}$

        T = 0.215 s

d) time when displacement is equal to x= 2.6 cm

x = (3.9 cm)sin[(9.3 rad/s)πt]

2.6 = (3.9 cm)sin[(9.3 rad/s)πt]

sin[(9.3 rad/s)πt] = 0.667

9.3 π t = 0.73

t = 0.025 s