Two masses $m_1=1 \,kg$ and $m_2=0.5 \,kg$ are suspended together by a massless spring of spring constant $12.5 \,Nm ^{-1}$. When masses are in equilibrium $m_1$ is removed without disturbing the system. New amplitude of oscillation will be .......... $cm$
Medium
Download our app for free and get started
(c)
Points of equilibrium of the spring will be when no force acts on it.
$k x=\left(m_1+m_2\right) g$
$x=\frac{\left(m_1+m_2\right) g}{k}$
The new equilibrium position which will be the mean position of $S.H.M.$ will be simply $\frac{m_2 g}{k}$
New amplitude will be maximum displacement from $\frac{m_2 g}{k}$ which is :
$A=\frac{\left(m_1+m_2\right) g}{k}-\frac{m_2 g}{k}$
or $A=\frac{m_1 g}{k}$
or $A=\frac{1 \times 10}{12.5}$
or $A=\frac{4}{5} \,m$
$\therefore A=0.8 \,m$ or $80 \,cm$
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
- 1Time period of a simple pendulum in a stationary lift is ' $T$ '. If the lift accelerates with $\frac{ g }{6}$ vertically upwards then the time period will be .....View Solution
(where $g =$ acceleration due to gravity)
- 2Initially system is in equilibrium. Time period of $SHM$ of block in vertical direction isView Solution
- 3A particle executes simple harmonic motion with an amplitude of $5\, cm$. When the particle is at $4\, cm$ from the mean position, the magnitude of its velocity is $SI\,units$ is equal to that of its acceleration. Then, its periodic time in seconds isView Solution
- 4A particle is performing simple harmonic motion with amplitude A and angular velocity ${\omega }$. The ratio of maximum velocity to maximum acceleration isView Solution
- 5A pendulum is suspended in a lift and its period of oscillation when the lift is stationary is $T_0$. What must be the acceleration of the lift for the period of oscillation of the pendulum to be $T_0/2$ ?View Solution
- 6A simple pendulum of frequency $f$ has a metal bob. If bob is charged negatively and is allowed to oscillate with large positive charged plate under it, frequency will beView Solution
- 7View SolutionTo make the frequency double of a spring oscillator, we have to
- 8A block is placed on a frictionless horizontal table. The mass of the block is m and springs are attached on either side with force constants ${K_1}$ and ${K_2}$. If the block is displaced a little and left to oscillate, then the angular frequency of oscillation will beView Solution
- 9A particle starts from a point $P$ at a distance of $A/2$ from the mean position $O$ and travels towards left as shown in the figure. If the time period of $SHM$ , executed about $O$ is $T$ and amplitude $A$ then the equation of motion of particle isView Solution
- 10A mass m performs oscillations of period $T$ when hanged by spring of force constant $K$. If spring is cut in two parts and arranged in parallel and same mass is oscillated by them, then the new time period will beView Solution
