The force constants of two springs are ${K_1}$ and ${K_2}$. Both are stretched till their elastic energies are equal. If the stretching forces are ${F_1}$ and ${F_2}$, then ${F_1}:{F_2}$ is
The force constants of two springs are ${K_1}$ and ${K_2}$. Both are stretched till their elastic energies are equal. If the stretching forces are ${F_1}$ and ${F_2}$, then ${F_1}:{F_2}$ is
- A
${K_1}:{K_2}$
- B
${K_2}:{K_1}$
- C
$\sqrt {{K_1}} :\sqrt {{K_2}} $
- D
$K_1^2:K_2^2$
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- [AIIMS 1998]
A particle of mass $m$ is attached to one end of a mass-less spring of force constant $k$, lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time $t=0$ with an initial velocity $u_0$. When the speed of the particle is $0.5 u_0$, it collies elastically with a rigid wall. After this collision :
$(A)$ the speed of the particle when it returns to its equilibrium position is $u_0$.
$(B)$ the time at which the particle passes through the equilibrium position for the first time is $t=\pi \sqrt{\frac{ m }{ k }}$.
$(C)$ the time at which the maximum compression of the spring occurs is $t =\frac{4 \pi}{3} \sqrt{\frac{ m }{ k }}$.
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A particle of mass $m$ is attached to one end of a mass-less spring of force constant $k$, lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time $t=0$ with an initial velocity $u_0$. When the speed of the particle is $0.5 u_0$, it collies elastically with a rigid wall. After this collision :
$(A)$ the speed of the particle when it returns to its equilibrium position is $u_0$.
$(B)$ the time at which the particle passes through the equilibrium position for the first time is $t=\pi \sqrt{\frac{ m }{ k }}$.
$(C)$ the time at which the maximum compression of the spring occurs is $t =\frac{4 \pi}{3} \sqrt{\frac{ m }{ k }}$.
$(D)$ the time at which the particle passes througout the equilibrium position for the second time is $t=\frac{5 \pi}{3} \sqrt{\frac{ m }{ k }}$.
- [IIT 2013]