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$1:3$
B$1:2$
C$2:3$
D$2:1$
Easy
Download our app for free and get started
D$2:1$
d (d) Force constant $(k)\, \propto \frac{1}{{{\rm{Length\, of\, the \,string \,(}}l{\rm{)}}}}$
==> $\frac{{{k_1}}}{{{k_2}}} = \frac{{{l_2}}}{{{l_1}}} = \frac{2}{1}$
Download our app
and get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
In an angular $SHM$ angular amplitude of oscillation is $\pi $ $rad$ and time period is $0.4\,sec$ then calculate its angular velocity at angular displacement $ \pi/2 \,rad$. ..... $rad/sec$
An object of mass $0.5\, {kg}$ is executing simple harmonic motion. Its amplitude is $5\, {cm}$ and time period (T) is $0.2\, {s} .$ What will be the potential energy of the object at an instant $t=\frac{T}{4}$ s starting from mean position. Assume that the initial phase of the oscillation is zero. (In ${J}$)
If the time period of a two meter long simple pendulum is $2\, s$, the acceleration due to gravity at the place where pendulum is executing $S.H.M.$ is
A particle of mass $250\,g$ executes a simple harmonic motion under a periodic force $F =(-25\,x) N$. The particle attains a maximum speed of $4\,m / s$ during its oscillation. The amplitude of the motion is $...........cm$.
A particle is executing simple harmonic motion $(SHM)$ of amplitude $A,$ along the $x-$ axis, about $x = 0.$ When its potential energy $(PE)$ equals kinetic energy $(KE),$ the position of the particle will be
Two particles $P$ and $Q$ describe $SHM$ of same amplitude $a$ , frequency $v$ along the same straight line. The maximum distance between the two particles is $a \sqrt 2$ . The initial phase difference between the particles is