A spring with $10$ coils has spring constant $k$. It is exactly cut into two halves, then each of these new springs will have a spring constant
A spring with $10$ coils has spring constant $k$. It is exactly cut into two halves, then each of these new springs will have a spring constant
- A
$k/2$
- B
$2k$
- C
$3k/2$
- D
$3k$
Similar Questions
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 weightless spring which has a force constant oscillates with frequency $n$ when a mass $m$ is suspended from it. The spring is cut into two equal halves and a mass $2m $ is suspended from it. The frequency of oscillation will now become
A weightless spring which has a force constant oscillates with frequency $n$ when a mass $m$ is suspended from it. The spring is cut into two equal halves and a mass $2m $ is suspended from it. The frequency of oscillation will now become
A mass m oscillates with simple harmonic motion with frequency $f = \frac{\omega }{{2\pi }}$ and amplitude A on a spring with constant $K$ , therefore
A mass m oscillates with simple harmonic motion with frequency $f = \frac{\omega }{{2\pi }}$ and amplitude A on a spring with constant $K$ , therefore
Initially system is in equilibrium. Time period of $SHM$ of block in vertical direction is
Initially system is in equilibrium. Time period of $SHM$ of block in vertical direction is
The mass $M$ shown in the figure oscillates in simple harmonic motion with amplitude $A$. The amplitude of the point $P$ is
The mass $M$ shown in the figure oscillates in simple harmonic motion with amplitude $A$. The amplitude of the point $P$ is
- [IIT 2009]