One end of a massless spring of spring constant k and natural length l0 is fixed while the other end is connected to a small object of mass m lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity ω about an axis passing through fixed end, then the elongation of the spring will be:

Question

One end of a massless spring of spring constant k and natural length l0 is fixed while the other end is connected to a small object of mass m lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity ω about an axis passing through fixed end, then the elongation of the spring will be: (A) ( k - m ω2 l0/ m ω2) (B) ( m ω2 l0/ k + m ω2) (C) ( m ω2 l0/ k - m ω2) (D) ( k + m ω2 l0/ m ω2)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/0bc0903ae3536165f1a41151f2817f6c-.png" /> K Δ x = m (ℓ0+ underline underlineΔ x ) w 2 K Δ x = m ℓ0 w 2+ mw 2 Δ x Δ x =( m ℓ0 w 2/ k - mw 2)

$\frac{k - m ω ^{2} l _{0}}{m ω ^{2}}$

$\frac{m ω ^{2} l _{0}}{k + m ω ^{2}}$

$\frac{m ω ^{2} l _{0}}{k - m ω ^{2}}$

$\frac{k + m ω ^{2} l _{0}}{m ω ^{2}}$

$K Δ x = m (ℓ_{0} + \underline{\underline{Δ}} x) w^{2}$
$K Δ x = m ℓ_{0} w^{2} + m w^{2} Δ x$
$Δ x = \frac{m ℓ _{0} w ^{2}}{k - m w ^{2}}$

mω2k−mω2l0​​

k+mω2mω2l0​​

k−mω2mω2l0​​

mω2k+mω2l0​​

KΔx=m(ℓ0​+Δ​​x)w2 KΔx=mℓ0​w2+mw2Δx Δx=k−mw2mℓ0​w2​

$\frac{k - m ω ^{2} l _{0}}{m ω ^{2}}$

$\frac{m ω ^{2} l _{0}}{k + m ω ^{2}}$

$\frac{m ω ^{2} l _{0}}{k - m ω ^{2}}$

$\frac{k + m ω ^{2} l _{0}}{m ω ^{2}}$

$K Δ x = m (ℓ_{0} + \underline{\underline{Δ}} x) w^{2}$
$K Δ x = m ℓ_{0} w^{2} + m w^{2} Δ x$
$Δ x = \frac{m ℓ _{0} w ^{2}}{k - m w ^{2}}$