Single Choice

A spring whose unstretched length is $$l$$ has a force constant $$k$$. The spring is cute into two pieces of unstretched lengths $$l_1$$ and $$l_2$$ where, $$l_1 = nl_2$$ and $$n$$ is an integer. The ratio $$\dfrac{k_1}{k_2}$$ of the corresponding force constants, $$k_1$$ and $$k_2$$ will be:

A$$\dfrac{1}{n^2}$$
B$$n^2$$
C$$\dfrac{1}{n}$$
Correct Answer
D$$n$$

Solution

$$k_1 = \dfrac{C}{\ell_1}$$

$$k_2 = \dfrac{C}{\ell_2}$$

$$\dfrac{k_1}{k_2} = \dfrac{C\ell_2}{\ell_1 C}, where, l_1=Cl_2\\\\ \dfrac{k_1}{k_2}= \dfrac{\ell_2}{n\ell_2} = \dfrac{1}{n}$$

SIMILAR QUESTIONS

Simple Harmonic Motion

Which of the following graph depicts spring constant $$K$$ versus length $$L$$ of the spring correctly?

Simple Harmonic Motion

A block of mass 0.5 kg hanging from a vertical spring executes simple harmonic motion of amplitude $$0.1 m$$ and time period $$0.314 s$$. Find the maximum force exerted by the spring on the block.

Simple Harmonic Motion

A $$5$$ kg collar is attached to a spring of spring constant $$500 Nm^{-1}$$. It slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by $$10$$cm and released. The time period of oscillation is:

Simple Harmonic Motion

An oscillator consists of a block attached to a spring ($$k= 400 N/m$$). At some time $$t$$, the position (measured from the system’s equilibrium location), velocity and acceleration of the block are $$x=0.100m$$, $$v=-13.6m/s$$, and $$a=-123m/s^{2}$$. Calculate (a) the frequency of oscillation, (b) the mass of the block and (c) the amplitude of the motion.

Simple Harmonic Motion

Above figure, shows block $$1$$ of mass $$0.200 kg$$ sliding to the right over a frictionless elevated surface at a speed of $$8.00 m/s$$. The block undergoes an elastic collision with stationary block $$2$$, which is attached to a spring of spring constant $$1208.5N/m$$. (Assume that the spring does not affect the collision.) After the collision, block $$2$$ oscillates in SHM with a period of $$0.140 s$$, and block $$1$$ slides off the opposite end of the elevated surface,landing a distanced from the base of that surface after falling height $$h = 4.90 m$$. What is the value of $$d$$?

Simple Harmonic Motion

The above figure gives the one dimensional potential energy well for a $$2.0kg$$ particle (the function $$U(x)$$ has the form $$bx^{2}$$ and the vertical axis scale is set by $$U_{s}=2.0J$$). (a) If the particle passes through the equilibrium position with a velocity of $$85cm/s$$, will it be turned back before it reaches $$x =15cm$$? (b) If yes, at what position, and if no, what is the speed of the particle at $$x=15cm$$?

Simple Harmonic Motion

A block sliding on a horizontal frictionless surface is attached to a horizontal spring with a spring constant of $$600 N/m$$. The block executes SHM about its equilibrium position with a period of $$0.40 s$$ and an amplitude of $$0.20 m$$. As the block slides through its equilibrium position, a $$0.50kg$$ putty wad is dropped vertically onto the block. If the putty wad sticks to the block, determine (a) the new period of the motion and (b) the new amplitude of the motion.

Simple Harmonic Motion

The vibration frequencies of atoms in solids at normal temperatures are of the order of $$10^{13}Hz$$. Imagine the atoms to be connected to one another by springs. Suppose that a single silver atom in a solid vibrates with this frequency and that all the other atoms are at rest. Compute the effective spring constant. One mole of silver ($$6.02*10^{23}$$ atoms) has a mass of $$108g$$.

Simple Harmonic Motion

A $$2\, kg$$ block moving with $$10\, m/s$$ strikes a spring of constant $$\pi ^2 N/m$$ attached to $$2\, kg$$ block at rest on a smooth floor. The time for which rear moving block remains in contact with spring will be

Simple Harmonic Motion

A block of mass $$m$$, attached to a fixed position $$O$$ on a smooth inclined wedge of mass $$M$$, oscillates with amplitude $$A$$ and linear frequency $$f$$. The wedge is located on a rough horizontal surface. If the angle of the wedge is $$60^o$$, then the force of friction acting on the wedge is given by ( coefficient of static friction $$=\mu$$):

Contact Details