In a seconds pendulum, mass of bob is $30\, gm$. If it is replaced by $90\, gm$ mass. Then its time period will .... $\sec$
Easy
Download our appand 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.*
Similar Questions
- 1When a particle of mass $m$ moves on the $x$-axis in a potential of the form $V(x)=\mathrm{kx}^2$ it performs simple harmonic motion. The corresponding time period is proportional to $\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$, as can be seen easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of $\mathrm{x}=0$ in a way different from $\mathrm{kx}^2$ and its total energy is such that the particle does not escape to infinity. Consider a particle of mass $\mathrm{m}$ moving on the $x$-axis. Its potential energy is $V(x)=\alpha x^4(\alpha>0)$ for $|x|$ near the origin and becomes a constant equal to $\mathrm{V}_0$ for $|x| \geq X_0$ (see figure). $Image$View Solution
$1.$ If the total energy of the particle is $E$, it will perform periodic motion only if
$(A)$ $E$ $<0$ $(B)$ $E$ $>0$ $(C)$ $\mathrm{V}_0 > \mathrm{E}>0$ $(D)$ $E > V_0$
$2.$ For periodic motion of small amplitude $\mathrm{A}$, the time period $\mathrm{T}$ of this particle is proportional to
$(A)$ $\mathrm{A} \sqrt{\frac{\mathrm{m}}{\alpha}}$ $(B)$ $\frac{1}{\mathrm{~A}} \sqrt{\frac{\mathrm{m}}{\alpha}}$ $(C)$ $\mathrm{A} \sqrt{\frac{\alpha}{\mathrm{m}}}$ $(D)$ $\mathrm{A} \sqrt{\frac{\alpha}{\mathrm{m}}}$
$3.$ The acceleration of this particle for $|\mathrm{x}|>\mathrm{X}_0$ is
$(A)$ proportional to $\mathrm{V}_0$
$(B)$ proportional to $\frac{\mathrm{V}_0}{\mathrm{mX}_0}$
$(C)$ proportional to $\sqrt{\frac{\mathrm{V}_0}{\mathrm{mX}_0}}$
$(D)$ zero
Give the answer qustion $1,2$ and $3.$
- 2A particle executes $SHM $ in a line $4\, cm $ long. Its velocity when passing through the centre of line is $12\, cm/s$. The period will be ..... $s$View Solution
- 3The length of the second pendulum on the surface of earth is $1\, m$. The length of seconds pendulum on the surface of moon, where g is 1/6th value of $g$ on the surface of earth, isView Solution
- 4A uniform rod of length $L$ and mass $M$ is pivoted at the centre. Its two ends are attached to two springs of equal spring constants $k$. The springs are fixed to rigid supports as shown in the figure, and the rod is free to oscillate in the horizontal plane. The rod is gently pushed through a small angle $\theta$ in one direction and released. The frequency of oscillation isView Solution
- 5Kinetic energy of a particle executing simple harmonic motion in straight line is $pv^2$ and potential energy is $qx^2$, where $v$ is speed at distance $x$ from the mean position. It time period is given by the expressionView Solution
- 6The amplitude of a particle executing $SHM$ is $4 \,cm$. At the mean position the speed of the particle is $16\, cm/sec$. The distance of the particle from the mean position at which the speed of the particle becomes $8\sqrt 3 \,cm/s,$ will be .... $cm$View Solution
- 7A mass $m$ attached to a spring oscillates with a period of $3\,s$. If the mass is increased by $1\,kg$ the period increases by $1\,s$. The initial mass $m$ isView Solution
- 8Figure shows the circular motion of a particle. The radius of the circle, the period, sense of revolution and the initial position are indicated in the figure. The simple harmonic motion of the $x-$ projection of the radius vector of the rotating particle $P$ isView Solution
- 9A ring is suspended from a point $S$ on its rim as shown in the figure. When displaced from equilibrium, it oscillates with time period of $1\,second.$ The radius of the ring is ..... $m$ (take $g = \pi ^2$ )View Solution
- 10Maximum amplitude(in $cm$) of $SHM$ so block A will not slip on block $B , K =100 N / m$View Solution
