Question

A particle performing SHM starts equilibrium position and its time period is 16 seconds. After 2 seconds its velocity is π m/s. Amplitude of oscillation is ______. `(cos 45° = 1/sqrt2)`

पर्याय

• 2`sqrt2` m

• 4`sqrt2` m

• 6`sqrt2` m

• 8`sqrt2` m

Answer 1

A particle performing SHM starts equilibrium position and its time period is 16 seconds. After 2 seconds its velocity is π m/s. Amplitude of oscillation is `underlinebb(8sqrt2  "m")`. `(cos 45° = 1/sqrt2)`

Explanation:

Given, particle velocity, v = πm/s and time period T = 16 s

Displacement of the particle, x = A sin ωt Velocity of the particle,

v = `("d"x)/"dt"` = Aω cos ωt                ...(i)

Angular velocity,

ω = `(2pi)/"T"=(2pi)/16=pi/8` = rad/s

Now from eq. (i),

π = `"A"xxpi/8xx"cos"pi/8xx2`

1 = `"A"/8"cos"pi/4="A"/8. 1/sqrt2`

∴ A = 8`sqrt2` m