Question

A body is performing simple harmonic with an amplitude of 10 cm. The velocity of the body was tripled by air Jet when it is at 5 cm from its mean position. The new amplitude of vibration is `sqrtx` cm. The value of x is ______.

विकल्प

• 800

• 600

• 700

• 500

Answer 1

A body is performing simple harmonic with an amplitude of 10 cm. The velocity of the body was tripled by air Jet when it is at 5 cm from its mean position. The new amplitude of vibration is `sqrtx` cm. The value of x is 700.

Explanation:

Let the equation of particle performing SHM be:

x = A sin (ωt)     …(i)

On differentiating equation (1) we get,

v = `"dx"/"dt"` = Aω cos ωt

⇒ v = aω `sqrt(1-sin^2 omega"t")`

 ⇒ v = Aω `sqrt(1- (x/"A")^2)`     ...[From eq (1)]

v = ω`sqrt("A"^2-x^2)`

Initially, the amplitude of SHM is 10 cm.

When the particle is at a location of 5 cm i.e., `"A"/2` it's speed is increased by air jet to 3V.

⇒ As a result, the SHM's overall energy will grow, increasing the amplitude as well.

⇒ Angular frequency i.e., w = `sqrt("Elastic Factor"/"Inertial factor")`

remains same because neither mass has changed nor any force.

v = ω `sqrt("A"^2-("A"/2)^2)`    ...`["When"  x = "A"/2]`...(1)

3v = `[omega sqrt(("A'")^2 - ("A"/2)^2 )  ] ["at"  x = "A"/2]`  ...(2)

Divide (1) by (2)

`1/3 = sqrt(("A"^2-"A"^2/4)/("A'"^2-"A"^2/4))`

On Sauaring both sides, we get;

`1/9 = (3"A"^2/4)/("A'"^2-"A"^2/4)`

⇒ `"A'"^2-"A"^2/4 = (27"A"^2)/4`

⇒ `"A'"^2 = (27/4+1/4) "A"^2/4 = 7"A"^2`

⇒ A' = `sqrt7"A"`

On comparison with `sqrtx`

x = 700 [As, A = 10 cm]