Answer 1

It is given that:
Maximum speed of the particle, \[v_{Max}\]= `10 "cm"^(- 1)`

Maximum acceleration of the particle,

\[a_{Max}\]= 50 cms⁻²​

The maximum velocity of a particle executing simple harmonic motion is given by,

\[v_{Max} = A\omega\]

where \[\text { omega is angular frequency, and }\]

A is amplitude of the particle.

Substituting the value of \[v_{Max}\]in the above expression,

we get :

Aω = 10     \[. . . (1)\]

\[\Rightarrow \omega^2 = \frac{100}{A^2}\] 

 

a_Max = ω²A = 50 cms⁻¹

\[\Rightarrow \omega^2 = \frac{50}{A} . . . (2)\]

\[\text { From the equations (1) and (2), we get:} \]

\[\frac{100}{A^2} = \frac{50}{A}\]

\[ \Rightarrow A = 2 cm\]

\[ \therefore \omega = \sqrt{\frac{100}{A^2}} = 5 \sec^{- 1}\]

To determine the positions where the speed of the particle is 8 ms^-1, we may use the following formula:
      v² = ω² (A² − y²)

where y is distance of particle from the mean position, and
                 v is velocity of the particle.

On substituting the given values in the above equation, we get:
      64 = 25 (4 − y²)

\[\Rightarrow \frac{64}{25} = 4 - y^2\]

⇒ 4 − y² = 2.56
⇒       y² = 1.44
⇒​       y  = \[\sqrt{1 . 44}\]

⇒        y = ± 1.2 cm   (from the mean position)