Physics · Oscillations and Simple Harmonic Motion
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A body performing SHM with displacement x=x 0sin(wt+Ⲫ),when t=0, x=x0..Then what is the phase angle of Ⲫ?
- A
π
- B
π/2
- C
π/4
- D
–π
Inside the sin bracket we need π/2 to get us a 1 in order to satisfy the condition which is x=x0. Since t=0, we would have wt as zero as well so the value of fi must be π/2.
This option is incorrect.
Inside the sin bracket we need π/2 to get us a 1 in order to satisfy the condition which is x=x0. Since t=0, we would have wt as zero as well
so the value of Ⲫ must be π/2.
According to the question, initially the body is at extreme position,
That means initial phase Ⲫ is 90 degree or 𝝅/2 rad.
Let’s prove it:
x=xosin(wt+Ⲫ)
x=xo, wt=0 degree and Ⲫ=90°
1=sin(90°
1=1 proved
This option is incorrect.
Inside the sin bracket we need π/2 to get us a 1 in order to satisfy the condition which is x=x0. Since t=0, we would have wt as zero as well
so the value of Ⲫ must be π/2.
This option is incorrect.
Inside the sin bracket we need π/2 to get us a 1 in order to satisfy the condition which is x=x0. Since t=0, we would have wt as zero as well
so the value of Ⲫ must be π/2.
Tagged under Physics · Oscillations and Simple Harmonic Motion · 2017