A particle in linear simple harmonic motion has a velocity of 4 ms-1 a

1.	A particle in linear simple harmonic motion has a velocity of 4 ms-1 at 3m at 3 ms-1 at 4 m from mean position. What is the time taken to travel half the amplitude from its positive extreme position?
Answer» Step 1. \(v=\omega\sqrt{a^2\,-\,y^2},\) \(v_2=\omega\sqrt{a^2\,-\,y_1^2}\) And \(v_2 =\omega\sqrt{a^2\,-y^2_2}\) 4 = \(\omega\sqrt{a^2\,-\,3^2}\) ...(i) And 3 = \(\omega\sqrt{a^2\,-\,4^2}\) ...(ii) Dividing eqn (i) by (ii) : \(\frac{4}{3}=\frac{\sqrt{a^2-9}}{\sqrt{a^2-16}}\) i.e., \(\frac{16}{9}=\frac{\sqrt{a^2-9}}{\sqrt{a^2-16}}\) Or 16a² − 256 = 9a² − 81 Or a²= 25 Or a = 5 m. Step 2. Putting a = 5 in eqn (i), 4 = 4ω Or ω = 1 rad s^-1 Step 3. Time taken to travel half amplitude from positive extreme positive extreme position is given by displacement, x =acos ωt i.e., \(\frac{5}{2}= 5cos(1\times t) \,or\,cos\, t=\frac{1}{2}\) Or t = cos^-1\(\frac{1}{2}\) = 60º = \(\frac{\pi}{3}\) Or t = \(\frac{3.142}{3}\) = 1.047 s.

A particle in linear simple harmonic motion has a velocity of 4 ms-1 at 3m at 3 ms-1 at 4 m from mean position. What is the time taken to travel half the amplitude from its positive extreme position?

Answer»

Step 1. \(v=\omega\sqrt{a^2\,-\,y^2},\)

\(v_2=\omega\sqrt{a^2\,-\,y_1^2}\)

And \(v_2 =\omega\sqrt{a^2\,-y^2_2}\)

4 = \(\omega\sqrt{a^2\,-\,3^2}\) ...(i)

And 3 = \(\omega\sqrt{a^2\,-\,4^2}\) ...(ii)

Dividing eqn (i) by (ii) :

\(\frac{4}{3}=\frac{\sqrt{a^2-9}}{\sqrt{a^2-16}}\)

i.e., \(\frac{16}{9}=\frac{\sqrt{a^2-9}}{\sqrt{a^2-16}}\)

Or 16a² − 256 = 9a² − 81

Or a²= 25

Or a = 5 m.

Step 2. Putting a = 5 in eqn (i),

4 = 4ω

Or ω = 1 rad s^-1

Step 3. Time taken to travel half amplitude from positive extreme positive extreme position is given by displacement,

x =acos ωt

i.e., \(\frac{5}{2}= 5cos(1\times t) \,or\,cos\, t=\frac{1}{2}\)

Or t = cos^-1\(\frac{1}{2}\) = 60º = \(\frac{\pi}{3}\)

Or t = \(\frac{3.142}{3}\) = 1.047 s.