If the maximum velocity and acceleration particle executing SHM are equal in magnitude the time period will be

Question

If the maximum velocity and acceleration particle executing SHM are equal in magnitude the time period will be (A) 1.57 s (B) 3.14 s (C) 6.28 s (D) 12.56 s

Tardigrade Admin · Accepted Answer

According to question | v max |=| a max | i.e., ω a=ω 2a ⇒ ω =1=(2π /T) ∴ T=2π ⇒ T=2× 3.14 ⇒ T=6.28 s

Q. If the maximum velocity and acceleration particle executing SHM are equal in magnitude the time period will be

1.57 s

3.14 s

6.28 s

12.56 s

According to question $∣ v_{m a x} ∣ = ∣ a_{m a x} ∣$ i.e., $ωa = ω^{2} a$ $\Rightarrow$ $ω = 1 = \frac{2 π}{T}$ $∴$ $T = 2 π$ $\Rightarrow$ $T = 2 \times 3.14$ $\Rightarrow$ $T = 6.28 s$

Q. If the maximum velocity and acceleration particle executing SHM are equal in magnitude the time period will be

1.57 s

3.14 s

6.28 s

12.56 s

According to question ∣vmax​∣=∣amax​∣ i.e., ωa=ω2a ⇒ ω=1=T2π​ ∴ T=2π ⇒ T=2×3.14 ⇒ T=6.28s

According to question $∣ v_{m a x} ∣ = ∣ a_{m a x} ∣$ i.e., $ωa = ω^{2} a$ $\Rightarrow$ $ω = 1 = \frac{2 π}{T}$ $∴$ $T = 2 π$ $\Rightarrow$ $T = 2 \times 3.14$ $\Rightarrow$ $T = 6.28 s$