Max and Min velocity of SHM

The maximum and minimum speed of a body executing simple harmonic motion with ω = 2 rad/s \omega =2 \text{ rad/s} and amplitude 5 5 meters is v max v_{\max} and v min v_{\min} respectively. Find v max + v min v_{\max} + v_{\min} (in meters per second).


The answer is 10.

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4 solutions

Sanjwal Singhs
Feb 19, 2016

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

y = A s i n ( ω t ) y=A sin(\omega t)

v = d y d t = A ω c o s ( ω t ) v=\frac{dy}{dt}=A \omega cos(\omega t)

v max = A ω v_{\max}= A \omega

since v min = 0 v_{\min}=0 ,

v t o t a l = v max = 5 × 2 = 10 v_{total}=v_{\max}=5 \times 2 =\boxed{10}

Aishwarya Dadhich
Feb 19, 2016

Easiest damn easy soty.
Ttttt

Vatsalya Tandon
Feb 1, 2016

Minimum Velocity is 0 0 at extreme postion or at y = A y=A where A A is the Amplitude.

Maximum Velocity is A ω = 10 m s A\omega = 10 \frac{m}{s} at mean postion or at y = 0 y=0 .

Therefore the sum is 10 m s 10 \frac{m}{s}

Minimum velocity could also be considered as 10 m s -10 \dfrac{m}{s} if you consider the directions. I think you should mention that.

A Former Brilliant Member - 5 years, 3 months ago

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