The maximum and minimum speed of a body executing simple harmonic motion with ω = 2 rad/s and amplitude 5 meters is v max and v min respectively. Find v max + v min (in meters per second).
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y = A s i n ( ω t )
v = d t d y = A ω c o s ( ω t )
v max = A ω
since v min = 0 ,
v t o t a l = v max = 5 × 2 = 1 0
Easiest damn easy soty.
Ttttt
Minimum Velocity is 0 at extreme postion or at y = A where A is the Amplitude.
Maximum Velocity is A ω = 1 0 s m at mean postion or at y = 0 .
Therefore the sum is 1 0 s m
Minimum velocity could also be considered as − 1 0 s m if you consider the directions. I think you should mention that.
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Let equation of shm be
X=a sin (wt) ---- (1) [w->omega or angular velocity] [x-> displacement of particle from mean position]
Also from 1 --- sin(wt) = x/a
Differentiating wrt time dX/dt = v = aw cos(wt)
Cos (y)= (1-sin^2(y))^1/2
So velocity(v) = aw (1-(x/a)^2)^1/2
Simplifying
V= w (a^2- x^2)^1/2
Clearly under root will never give -ve value So max is when x=o (mean position) ie aw And min is when x=a (extreme position) ie 0
Vmax + Vmin = aw = 2 * 5 = 10