A particle performing SHM starts equilibrium position and its time period is 16 seconds. After 2 seconds its velocity is `pi m//s`. Amplitude of oscil

Question

A particle performing SHM starts equilibrium position and its time period is 16 seconds. After 2 seconds its velocity is `pi m//s`. Amplitude of oscil

A particle performing SHM starts equilibrium position and its time period is 16 seconds. After 2 seconds its velocity is `pi m//s`. Amplitude of oscillation is
`(cos 45^(@) = 1/(sqrt(2)))`
A. `2sqrt2m`
B. `4sqrt2m`
C. `6sqrt2m`
D. `8sqrt2m`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - D
Displacement of the particle `=x=A sin omegat`
Velocity of the particle `=v=(dx)/(dt)=A omega cos omegat`
`v=pi m//s, T=16s, omega=(2pi)/(T)=(2pi)/(16)"rad/s"`
`therefore pi=A xx (pi)/(8)xx cos""(pi)/(8)xx2`
`therefore 1=(A)/(8)cos ""pi/4=(A)/(8).(1)/(sqrt2)`
`therefore A=8sqrt2m`

4 views Answered 2023-02-05 15:29