Tag: horizontal oscillations of a mass attached to a spring

Questions Related to horizontal oscillations of a mass attached to a spring

A body of mass 'm' when hung from  a long and light spring, the spring stretches by 20 cm The period of vibration of the mass when pulled down the released is 

horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

  1. $\dfrac{2\pi}{7}$

  2. $4$

  3. $ \pi$

  4. $\dfrac{\pi}{\sqrt 3}$

Show answer
Correct Option: A

A sphere of mass 1 kg is connected to a spring of spring constant $ 5.0 Nm^{-1} $ as shown in figure. A force of 0.5 N is applied on the sphere along X-axis , what is the velocity of the sphere when it is displaced througha distance of 10 cm along X-axis?

horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

  1. $ 0.11 ms^{-1} $

  2. $ 0.22 ms^{-1} $

  3. $ 0.11 cms^{-1} $

  4. $ 0.22 cms^{-1} $

Show answer
Correct Option: C

A spring of length $'l'$ has spring constant $'k'$ is cut into two parts of length $l _{1}$ and $l _{2}$. If their respective spring constants are $k _{1}$ and $k _{2}$, then $\dfrac {k _{1}}{k _{2}}$ is

horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

  1. $\dfrac {l _{2}}{l _{1}}$

  2. $\dfrac {2l _{2}}{l _{1}}$

  3. $\dfrac {l _{1}}{l _{2}}$

  4. None

Show answer
Correct Option: A
Explanation:

$k _{1}l _{1} = k _{2}l _{2} = kl$
$\dfrac {k _{1}}{k _{2}} = \dfrac {l _{2}}{l _{1}}$

A block falls from a table $0.6m$ high. It lands on an ideal, mass-less, vertical spring with a force constant of $2.4kN/m$. The spring is initially $25cm$ high, but it is compressed to a minimum height of $10cm$ before the block is stopped. Find the mass of the block $(g=9.81m/s^2)$.

horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

  1. $55.51kg$

  2. $5.51kg$

  3. $0.51kg$

  4. None

Show answer
Correct Option: B

A block of mass $m=4$ kg undergoes simple harmonic motion with amplitude $A=6$ cm on the frictionless surface. Block is attached to a spring of force constant $k=400 N/m$. If the block is at $x = 6$ cm at time $t = 0$ and equilibrium position is at $x=0$ then the blocks position as a function of time (with $x$ in centimetres and $t$ in seconds)?

horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

  1. $x=6\,sin(10t+\frac{1}{2}\pi)$

  2. $x=6\,sin(10\pi t)$

  3. $x=6\,sin(10\pi t-\frac{1}{2}\pi)$

  4. $x=6\,sin(10t-\frac{1}{4}\pi)$

Show answer
Correct Option: A
Explanation:


The position of block executing SHM is $x=A\sin\left(\dfrac{{2}{\pi}t}{T}+{\phi}\right)$

where $\phi$ is the initial phase 

given at $t=0$    , $x=6cm$ and $A=6cm$

therefore by above equation $6=6\sin\left(0+{\phi}\right)$

                                        or  $1=\sin\phi$

                                        or $\sin\dfrac{\pi}{2}=\sin\phi$

                                        or $\phi=\dfrac{\pi}{2}$

now time period of system $T=2\pi\sqrt\frac{m}{k}$

                                          $T=2\pi\sqrt\frac{4}{400}$

                                          $T=\pi/5$

so position of block $x=6\sin\left(10t+\dfrac{\pi}{2}\right)$

as the initial phase is in positive x-direction therefore $+ $ sign is taken there.

When a spring-mass system vibrates with simple harmonic motion, the mass in motion reaches its maximum velocity:

horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

  1. when its acceleration is greatest

  2. when its acceleration is least.

  3. once during one oscillation.

  4. when it is as its maximum displacement from equilibrium.

  5. when it experiences maximum force.

Show answer
Correct Option: B
Explanation:

Velocity in SHM is given by ,

                              $v=\omega\sqrt{a^{2}-y^{2}}$ ,
where $\omega=$ angular velocity ,
           $a=$ amplitude ,
           $y=$ displacement from mean position ,
velocity $v$ will be maximum when RHS of this relation is maximum , this is maximum when $y$ is minimum i.e.$y=0$ ,
now acceleration in SHM is given by ,
                                $A=-\omega^{2}y$ ,
therefore acceleration at $y=0$ will be ,
                                $A=0$ ,
it implies that velocity is maximum when acceleration is least .

A mass of $36$ kg is kept vertically on the top of a massless spring. What is the maximum compression of the spring if the spring constant is $15000 $N/m. Assume $g=10 m/s^2$.
horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

  1. 0.02 m

  2. 0.14m

  3. 0.0004m

  4. 0.2m

  5. 42.52m

Show answer
Correct Option: A
Explanation:

When the maximum compression of the spring occurs, the forces of gravity and that from spring due to compression balance each other.

At that point, $mg=kx$
$\implies x=\dfrac{mg}{k}$
$=\dfrac{36\times 10}{15000}$
$\approx 0.02m$

A block is attached to an ideal spring undergoes simple harmonic oscillations of amplitude A. Maximum speed of block is calculated at the end of the spring. If the block is replaced by one with twice the mass but the amplitude of its oscillations remains the same, then the maximum speed of the block will

horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

  1. decrease by a factor of 4

  2. decrease by a factor of 2

  3. decrease by a factor of $\sqrt 2$

  4. remain the same

  5. increase by a factor of 2

Show answer
Correct Option: C
Explanation:

Maximum speed of the block         $V _{max}    = wA$                     where  $w  = \sqrt{\dfrac{k}{m}}$

$\implies$    $V _{max}   = \sqrt{\dfrac{k}{m}} A$                        where $k  = constant$

Now the mass is doubled  i.e  $m' = 2m$  keeping  $A  =constant$ 
$\therefore$       $V' _{max} = \sqrt{\dfrac{k}{2m}} A   = \dfrac{  V _{max}}{\sqrt{2}}$
Thus speed is decreased by a factor of  $\sqrt{2}$.

A block of mass $m$ attached to an ideal spring undergoes simple harmonic motion. The acceleration of the block has its maximum magnitude at the point where :

horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

  1. the speed is the maximum

  2. the potential energy is the minimum

  3. the speed is the minimum

  4. the restoring force is the minimum

  5. the kinetic energy is the maximum

Show answer
Correct Option: C
Explanation:

The magnitude of acceleration of the simple harmonic motion is $a=F/m=kx/m$

So, the acceleration is maximum when the displacement $x$ from equilibrium will be maximum. Thus, speed will be minimum and also the potential energy will also be maximum and kinetic energy will be minimum at the end points of the oscillation.  

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 = 13.6 m/s, and a = 123 m/s$^2$. The amplitude of the motion and the mass of the block are

horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

  1. $0.2 m, 0.845 kg$

  2. $0.3 m, 0.765 kg$

  3. $0.4 m, 0.325 kg$

  4. $0.5 m, 0.445 kg$

Show answer
Correct Option: D