Question

Two colinear simple harmonic motions acting simultaneously on a particle are given by x=0.3cos2Πt and y=0.2 sin(2Πt-Π/3) . where x is expressed in cm and t in seconds. Write down the expression for the resultant displacement as a function of time. ​

Answer 1

Answer:

y=0.3sin20π(t+0.05)

y=0.3sin(20πt+π)

Comparing the given equation with the standard equation y=Asin(ωt+ϕ)

By comparison

ω=20π

⇒phi=π

⇒A=0.3

Thus, amplitude and initial phase is known

Amplitude (A)=0.3m

Initial phase (ϕ)=π

Then, for calculating time period

T=

ω

2π

=

20π

2π

=

10

1

s=0.1s

⇒ Time period (T)=0.1s

Also for initial displacement t=0

⇒y=0.3sin(π)=0

Thus, initial displacement =0.

Answer 2

Explanation:

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