If a spring extends by $x$ on loading, then energy stored by the spring is (if $T$ is the tension in the spring and $K$ is the spring constant)
If a spring extends by $x$ on loading, then energy stored by the spring is (if $T$ is the tension in the spring and $K$ is the spring constant)
- A
$\frac{{{T^2}}}{{2x}}$
- B
$\frac{{{T^2}}}{{2K}}$
- C
$\frac{{2K}}{{{T^2}}}$
- D
$\frac{{2{T^2}}}{K}$
Similar Questions
A uniform spring of force constant $k$ is cut into two pieces, the lengths of which are in the ratio $1 : 2$. The ratio of the force constants of the shorter and the longer pieces is
A uniform spring of force constant $k$ is cut into two pieces, the lengths of which are in the ratio $1 : 2$. The ratio of the force constants of the shorter and the longer pieces is
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 $t=\frac{5 \pi}{3} \sqrt{\frac{ m }{ k }}$.
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]
A $5\; kg$ collar is attached to a spring of spring constant $500\;N m ^{-1} .$ It slides without friction over a hortzontal rod. The collar is displaced from its equilibrium position by $10.0\; cm$ and released. Calculate
$(a)$ the period of oscillation.
$(b)$ the maximum speed and
$(c)$ maximum acceleration of the collar.
A $5\; kg$ collar is attached to a spring of spring constant $500\;N m ^{-1} .$ It slides without friction over a hortzontal rod. The collar is displaced from its equilibrium position by $10.0\; cm$ and released. Calculate
$(a)$ the period of oscillation.
$(b)$ the maximum speed and
$(c)$ maximum acceleration of the collar.
A mass m is suspended from a spring of length l and force constant $K$. The frequency of vibration of the mass is ${f_1}$. The spring is cut into two equal parts and the same mass is suspended from one of the parts. The new frequency of vibration of mass is ${f_2}$. Which of the following relations between the frequencies is correct
A mass m is suspended from a spring of length l and force constant $K$. The frequency of vibration of the mass is ${f_1}$. The spring is cut into two equal parts and the same mass is suspended from one of the parts. The new frequency of vibration of mass is ${f_2}$. Which of the following relations between the frequencies is correct
How the period of oscillation depend on the mass of block attached to the end of spring ?
How the period of oscillation depend on the mass of block attached to the end of spring ?