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 =
[tex]f = \dfrac{\omega}{2\pi}[/tex]
[tex]f = \dfrac{9.3\pi}{2\pi}[/tex]
f = 4.65 Hz
b) period of motion
[tex]T= \dfrac{1}{f}[/tex]
[tex]T= \dfrac{1}{4.65}[/tex]
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
The frequency of the wave is 4.65 Hz.
The period of the motion is 0.22 s.
The amplitude of the wave is 3.9 cm.
The time when the object reaches 2.6 cm is 0.025 s.
The given parameters;
wave equation, [tex]x = 3.9 cm\ sin[(9.3 \ rad/s)\pi t][/tex]
The frequency of the wave is calculated as follows;
[tex]\omega = 2\pi f\\\\f = \frac{\omega}{2\pi} \\\\f = \frac{9.3\pi }{2\pi} \\\\f = 4.65 \ Hz[/tex]
The period of the motion is calculated as follows;
[tex]T = \frac{1}{f} = \frac{1}{4.65} = 0.22 \ s[/tex]
The amplitude of the wave is obtained by comparing the given equation with general wave equation;
Amplitude, A = 3.9 cm
The time when the object reaches 2.6 cm is calculated as follows;
[tex]2.6= 3.9 sin[(9.30 ) \pi t]\\\\\frac{2.6}{3.9} = sin[(9.3)\pi t]\\\\0.667 = sin[(9.3)\pi t]\\\\sin^{-1}(0.667) = 9.3 \pi t \\\\0.73 = 9.3 \pi t\\\\t = \frac{0.73}{9.3 \pi} \\\\t = 0.025 \ s[/tex]
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