The variation of velocity of a particle executing SHM with time is shown in figure. the maximum displacement and acceleration of the particle are respectively:

Question

The variation of velocity of a particle executing SHM with time is shown in figure. the maximum displacement and acceleration of the particle are respectively: (A) (10/π) m and 5 π m / s 2 (B) (5/π) m and (5 π/2) m / s 2 (C) (10/π) m and (5 π/2) m / s 2 (D) (5/π) m and (5 π/4) m / s 2

Tardigrade Admin · Accepted Answer

Maximum displacement, x max =A=(10/π) ⋅ m and maximum acceleration, a max =ω2 A =((π/2))2 × (10/π)=(5 π/2) m / s 2

Q. The variation of velocity of a particle executing SHM with time is shown in figure. the maximum displacement and acceleration of the particle are respectively:

$\frac{10}{π} m$ and $5 πm / s^{2}$

$\frac{5}{π} m$ and $\frac{5 π}{2} m / s^{2}$

$\frac{10}{π} m$ and $\frac{5 π}{2} m / s^{2}$

$\frac{5}{π} m$ and $\frac{5 π}{4} m / s^{2}$

Maximum displacement, $x_{m a x} = A = \frac{10}{π} \cdot m$
and maximum acceleration, $a_{m a x} = ω^{2} A$
$= (\frac{π}{2})^{2} \times \frac{10}{π} = \frac{5 π}{2} m / s^{2}$

Q. The variation of velocity of a particle executing SHM with time is shown in figure. the maximum displacement and acceleration of the particle are respectively:

π10​m and 5πm/s2

π5​m and 25π​m/s2

π10​m and 25π​m/s2

π5​m and 45π​m/s2

Maximum displacement, xmax​=A=π10​⋅m and maximum acceleration, amax​=ω2A =(2π​)2×π10​=25π​m/s2

$\frac{10}{π} m$ and $5 πm / s^{2}$

$\frac{5}{π} m$ and $\frac{5 π}{2} m / s^{2}$

$\frac{10}{π} m$ and $\frac{5 π}{2} m / s^{2}$

$\frac{5}{π} m$ and $\frac{5 π}{4} m / s^{2}$

Maximum displacement, $x_{m a x} = A = \frac{10}{π} \cdot m$
and maximum acceleration, $a_{m a x} = ω^{2} A$
$= (\frac{π}{2})^{2} \times \frac{10}{π} = \frac{5 π}{2} m / s^{2}$