$A$ spring with a spring constant $1200 \; N m^{-1}$ is mounted on a horizontal table as shown in the figure. $A$ mass of $3 \; kg$ is attached to the free end of the spring. The mass is then pulled sideways to a distance of $2.0 \; cm$ and released. Determine:
$(i)$ the frequency of oscillations,
$(ii)$ maximum acceleration of the mass,and
$(iii)$ the maximum speed of the mass.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) Given:
Spring constant,$k = 1200 \; N m^{-1}$
Mass,$m = 3 \; kg$
Amplitude (displacement),$A = 2.0 \; cm = 0.02 \; m$
$(i)$ Frequency of oscillation $(v)$:
The frequency is given by $v = \frac{1}{2 \pi} \sqrt{\frac{k}{m}}$
$v = \frac{1}{2 \times 3.14} \sqrt{\frac{1200}{3}} = \frac{1}{6.28} \sqrt{400} = \frac{20}{6.28} \approx 3.18 \; Hz$
$(ii)$ Maximum acceleration $(a_{max})$:
The maximum acceleration is given by $a_{max} = \omega^2 A$,where $\omega = \sqrt{\frac{k}{m}}$
$a_{max} = \frac{k}{m} A = \frac{1200}{3} \times 0.02 = 400 \times 0.02 = 8.0 \; m s^{-2}$
$(iii)$ Maximum speed $(v_{max})$:
The maximum speed is given by $v_{max} = A \omega$
$v_{max} = A \sqrt{\frac{k}{m}} = 0.02 \times \sqrt{\frac{1200}{3}} = 0.02 \times 20 = 0.4 \; m s^{-1}$

Explore More

Similar Questions

Find the maximum amplitude for safe $SHM$ (the block does not topple during $SHM$) of a cubical block of side $a$ on a smooth horizontal floor as shown in the figure (the spring is massless). The spring is attached at a height of $2a/3$ from the floor.

Difficult
View Solution

When a mass $m$ is attached to a spring,it normally extends by $0.2\, m$. If the mass $m$ is given a slight additional extension and released,what will be its time period?

$A$ particle of mass $m$ is attached to three identical springs $A, B$ and $C$ each of force constant $k$ as shown in the figure. If the particle of mass $m$ is pushed slightly against the spring $A$ and released,then the time period of oscillations is

Difficult
View Solution

The motion of a mass on a spring,with spring constant $K$ is as shown in the figure. The equation of motion is given by $x(t) = A \sin \omega t + B \cos \omega t$ with $\omega = \sqrt{\frac{K}{m}}$. Suppose that at time $t = 0$,the position of the mass is $x(0)$ and velocity is $v(0)$,then its displacement can also be represented as $x(t) = C \cos (\omega t - \phi)$,where $C$ and $\phi$ are:

The angular frequency of a spring-block system is $\omega_0$. This system is suspended from the ceiling of an elevator moving downwards with a constant speed $v_0$. The block is at rest relative to the elevator. The lift is suddenly stopped. Assuming the downward direction as positive,choose the wrong statement:

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo