Tag: simple harmonic motion

Questions Related to simple harmonic motion

The particle executes SHM on a straight line. At two positions its velocity $u$ and $v$ while acceleration, $\alpha$ and $\beta$ respectively $[\beta > \alpha >0]$, the distance between the two positions will be:-

  1. $\frac{u^2+v^2}{\alpha+\beta}$

  2. $-\frac{u^2-v^2}{\alpha+\beta}$

  3. $\frac{u^2-v^2}{\alpha-\beta}$

  4. $\frac{u^2+v^2}{\beta-\alpha}$


Correct Option: B
Explanation:
$u=w\sqrt{A^2-X _1^2}$
$V=w\sqrt{A^2-X _2^2}$
$\alpha=-w^2 x _1^2$
$\beta=-w^2 x _2^2$
$\frac{u^2-v^2}{u^2}=-(x _1^2-x _2^2)$
$=-(x _1+x _2)(x _1-x _2)$
$=\frac{(\alpha+\beta)}{w^2}(x _1-x _2)$
$x _1-x _2=\frac{u^2-v^2}{\alpha+\beta}$

A stone is projected from the ground with a velocity of $14 \ ms^{-1}$ one second later it clean a wall $2 \ m$ high. The angle of projection is $(g = 10 \ ms^{-2})$ 

  1. $30 ^\circ$

  2. $45 ^\circ$

  3. $60 ^\circ$

  4. $15 ^\circ$


Correct Option: A

The period of a particle it is $8s$. At $t=0$ it is at the mean position. The ratio of the distance covered by the particle in first second and second will be

  1. $\cfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } } $

  2. $\cfrac { 1 }{ \sqrt { 2 } } $

  3. $\cfrac { 1 }{ \sqrt { 2 } -1 } $

  4. $\left[ \sqrt { 2 } -1 \right] $


Correct Option: C
Explanation:

$x=A\sin\omega t$

$T= 8 sec$
$\dfrac{2\pi}{\omega}=8 sec$
$\omega=\dfrac{2\pi}{8}=\dfrac{\pi}{4}$
$x=A\sin \dfrac{\pi}{4} t$ ,At $t=1 sec$
$x _1=\dfrac{A}{\sqrt2}$ ,At $t=2sec$
$x _2=A$
Required ration $=\dfrac{x _1}{x _2-x _1}=\dfrac{1/\sqrt2}{1-\dfrac{1}{\sqrt2}}=\dfrac{1}{\sqrt2-1}$

Sunlight of intensity 1.3 kW m–2 is incident normally on a thin convex lens of focal length 20 cm. Ignore the energy loss of light due to the lens and assume that the lens aperture size is much smaller than its focal length. The average intensity of light, kW m–2, at a distance 22 cm from the lens on the other side is _________. 

  1. $ 130 $

  2. $120 \ $

  3. $ 170$

  4. $160 $


Correct Option: A,C

Simple harmonic motion is the projection of uniform circular motion on the

  1. $x$- axis

  2. $y$- axis

  3. reference circle

  4. any diameter of reference circle.


Correct Option: D
Explanation:

Simple harmonic motion is the projection of uniform circular motion on any diameter of reference circle.

A body of mass 'm' is suspended with an ideal spring of force constant 'k'. The expected change in the position of the body, due to an additional force 'F' acting vertically downwards is 

  1. $\cfrac { 3F }{ 2K } $

  2. $\cfrac { 2F }{ K } $

  3. $\cfrac { 5F }{ 2K } $

  4. $\cfrac { 4F }{ K } $


Correct Option: B

A mass of 2 kg falls from a height of 40 cm, on a spring with a force constant of 1960 N/m. The spring is compressed by ? (Take $g=9.8m/s^2$)

  1. 9 cm

  2. 1.0 cm

  3. 20 cm

  4. 5 cm


Correct Option: A

A string fixed at both ends vibrates in a resonant mode with separation of $6.0$cm between the consecutive nodes. For the next to next higher resonant frequency this separation is reduced by $2.0$cm. The length of the spring is 

  1. 8 cm

  2. 16 cm

  3. 24 cm

  4. 32 cm


Correct Option: A

A man weighing 60 kg stands on the horizontal platform of a spring balance. The platform starts executing simple harmonic motion of amplitude 0.1 m and frequency $2/ \pi$ Hz. Which of the following statements is correct?

  1. The spring balance reads the weight of man as 60kg

  2. The spring balance reading fluctuates between 60 kg. and 70 kg

  3. The spring balance reading fluctuates between 50 kg and 60 kg

  4. The spring balance reading fluctuates between 50 kg and 70 kg


Correct Option: A
Explanation:

Here,

The option $A$ is the correct answer because

as you know spring balance observe normal reaction between contacting surface. it is effected only when lift accelerated or decelerated . 
if lift is moving upward with acceleration $a $
then, observation of spring balance will be $= m(g + a)$ , where m is mass of man 
when lift is moving downward with acceleration a then, observation of spring balance will be $= m(g - a) .$
but when lift is moving upward or donward with constant velocity then, observation will be remain same.
hence, observation of man's weight is $60kg$ on spring balance.

A loaded spring gun. Initially at rest on a horizontal frictioneles surface fires a marble of  mass m in at an angle of elevation ${ 0 }^{ o }$. The mass of the gun is M that of the marble is m and its muzzle velocity of the marble is ${ V } _{ 0 }$ then Velocity of the gem just after the firing is 

  1. $\dfrac { m{ v } _{ 0 } }{ M } $

  2. $\dfrac { m{ v } _{ 0 }\cos { \theta } }{ M } $

  3. $\dfrac { m{ v } _{ 0 }\cos { \theta } }{ M+m } $

  4. $\dfrac { m{ v } _{ 0 }\cos { 2\theta } }{ M+m } $


Correct Option: A