Q: The amplitude of a particle performing S.H.M. is \$A\$. The displacement at which its velocity will be half of the maximum velocity is

\$x = A sin \omega t\$\$V=\dfrac{dx}{dt}=A\omega cos \omega t\$\$V_{max} =A \omega\$\$V=\dfrac{V_{max}}{2}= A \omega cos \omega t\$\$\Rightarrow \dfrac{A \omega }{2}=A \omega cos \omega t\$\$\Rightarrow cos \omega t=\dfrac{1}{2}\$\$\Rightarrow \omega t =60^{0}\$\$x=A sin \omega t\$\$=A sin 60^{0}\$\$=A \dfrac{\sqrt{3}}{2}\$so, amplitude is \$\dfrac{\sqrt{3}A}{2}\$

Q: The equation of the displacement of two particles making SHM are represented by \$ y_{1} \$ = a sin \$ \left ( \omega t +\phi \right ) \$ & \$ y_{2} \$ = a cos \$ \left ( \omega t \right ) \$. The phase difference of the velocities of the two particles is :

\$y_{1}=a sin (\omega t+\phi )\$\$y_{2}=a cos (\omega t)\$\$y_{1}=a \omega cos (\omega t+\phi)=a \omega sin (\omega t+\phi +{\pi}/_{2})\$\$y_{2}=-a \omega sin (\omega t)= a \omega sin (\pi + \omega t)\$therefore phase difference\$ = \omega t+\phi +{\pi}/_{2}-(\pi+\omega t)\$\$=\phi -{\pi}/_{2}\$

Q: Acceleration-displacement graph of a particle executing SHM is as shown in the figure. The time period of oscillation is (in sec)

In SHM\$a=-\omega^2x\$or\$a= \dfrac{-K}{m}x\$so, from graph\$-\dfrac{K}{m}= -1\$ \$ ( \because slope\ is\ -1)\$\$\dfrac{K}{m}= 1\$Time period \$= 2\pi \sqrt{\dfrac{m}{K}}\$\$= 2\pi \sqrt{\dfrac{1}{1}}\$\$= 2\pi \$

Q: \$ T _{1}\$, \$ T _{2}\$ are time periods of oscillation of a block individually suspended to spring of force constants \$ K _{1}\$ ,\$ K _{2}\$ respectively. If same block is suspended to parallel combination of same two springs, its time period is

for a parallel combination \$k=k_{1}+k_{2}\$\$T=2\pi \sqrt{\dfrac{m}{k}}\$ \$=2\pi \sqrt{\dfrac{m}{(k_{1}+k_{2})}}\$ \$T_{1}=2\pi \sqrt{\dfrac{m}{k_{1}}}\$ \$\Rightarrow k_{1}=(2\pi )^{2}\dfrac{m}{T_{1}^{2}}\$ \$k_{2}=(2\pi )^{2}\dfrac{m}{T_{2}^{2}}\$ \$k_{1}+k_{2}=(2\pi )^{2}(\dfrac{m}{T_{1}^{2}}+\dfrac{m}{T_{2}^{2}})\$ \$\therefore \dfrac{T_{1}T_{2}}{\sqrt{T_{1}+T_{2}}}=T\$

Q: An object is attached to the bottom of a light vertical spring and set vibrating. The maximum speed of the object is \$15 cm/sec\$ and the period is \$628\$ milliseconds. The amplitude of the motion in centimeters is:

\$v_{max}=15\ cm /s ; T=0.628\ s=\dfrac{2\pi}{10}s\$\$\omega =\dfrac{2\pi }{T}=10\ rad/s\$\$v_{max}=A\omega \$\$A=1.5\ cm\$

Q: A man of mass 60 kg, standing on a platform, is executing SHM in a vertical plane. The displacement from mean position is \$y = 0.5 sin (2\pi t)\$. The minimum value of frequency (\$f\$) for which the man will feel weightlessness at the highest point is :

\$w=2\pif\$\$A=0.5\$\$a=-w^{2}A\$For weightlessness we have \$a=g\$Thus\$g=-w^{2}A\$\$w^{2}=\dfrac{g}{0.5}=2g\$\$4\pi^{2}f^{2}=2g\$\$f^{2}=\dfrac{g}{2\pi^{2}};f=\dfrac{\sqrt{2g}}{2\pi}\$

Q: If the displacement \$x\$ and velocity \$v\$ of a particle executing S.H.M are related through the expression \$ 4v^{2}=25-x^{2} \$, then its maximum displacement in meters is

\$v=\omega\sqrt{A^{2}-x^{2}}\$\$4v^{2}=5^{2} -x^{2} \implies v=\frac{1}{2} \sqrt{(5^{2}-x^{2})}\$Comparing the equation with the standard equation we get \$A = \$5m

Q: The displacement equations of two simple harmonic oscillators are given by \$ x_{1} =A_{1}\cos\omega t\$; \$ x_{2}= A_{2} \sin\left(\omega t+\frac{\pi }{6} \right )\$. The phase difference between them is:

\$X_{1}=A_{1} cos\omega t =A_{1} sin (\omega t+^{\pi}/_{2})\$\$X_{2}=A_{2} sin (\omega t+^{\pi}/_{6})\$so, phase difference \$=\dfrac{\pi}{2}-\dfrac{\pi}{6}=\dfrac{\pi}{3}=60^{0}\$

Q: The amplitude of oscillation of a particle is 0.05 m.If its period is 1.57s. Then the velocity at the mean position is

\$A=0.05m\$\$T=\dfrac{\pi }{2}sec ; w=4\ { rad}/{sec}\$\$v_{max}=wA =0.2 \ {m}/{s}\$

Question 1

The amplitude of a particle performing S.H.M. is $A$. The displacement at which its velocity will be half of the maximum velocity is

Accepted Answer

$x = A sin \omega t$$V=\dfrac{dx}{dt}=A\omega cos \omega t$$V_{max} =A  \omega$$V=\dfrac{V_{max}}{2}= A \omega  cos \omega t$$\Rightarrow \dfrac{A  \omega }{2}=A  \omega cos \omega t$$\Rightarrow cos \omega t=\dfrac{1}{2}$$\Rightarrow \omega t =60^{0}$$x=A sin \omega t$$=A  sin  60^{0}$$=A \dfrac{\sqrt{3}}{2}$so, amplitude is $\dfrac{\sqrt{3}A}{2}$

Question 2

The equation of the displacement of two particles making SHM are represented by $ y_{1} $ = a sin $ \left ( \omega t +\phi \right ) $ & $ y_{2} $ = a cos $ \left ( \omega t \right ) $. The phase difference of the velocities of the two particles is :

Accepted Answer

$y_{1}=a    sin (\omega t+\phi  )$$y_{2}=a   cos (\omega t)$$y_{1}=a   \omega cos (\omega t+\phi)=a \omega sin (\omega t+\phi +{\pi}/_{2})$$y_{2}=-a   \omega sin (\omega t)= a \omega sin (\pi + \omega t)$therefore phase difference$ = \omega t+\phi +{\pi}/_{2}-(\pi+\omega t)$$=\phi -{\pi}/_{2}$

Question 3

Acceleration-displacement graph of a particle executing SHM is as shown in the figure. The time period of oscillation is (in sec)

Accepted Answer

In SHM$a=-\omega^2x$or$a= \dfrac{-K}{m}x$so, from graph$-\dfrac{K}{m}= -1$                   $ ( \because slope\ is\ -1)$$\dfrac{K}{m}= 1$Time period $= 2\pi \sqrt{\dfrac{m}{K}}$$= 2\pi \sqrt{\dfrac{1}{1}}$$= 2\pi $

Question 4

$ T _{1}$, $ T _{2}$ are time periods of oscillation of a block individually suspended to spring of force constants $ K _{1}$ ,$ K _{2}$ respectively. If same block is suspended to parallel combination of same two springs, its time period is

Accepted Answer

&#10;&#10;&#10;&#9;&#10;&#9;&#10;&#9;&#10;&#9;&#10;&#10;&#10;&#10;&#10;for a parallel&#10;combination $k=k_{1}+k_{2}$$T=2\pi \sqrt{\dfrac{m}{k}}$&#10;$=2\pi&#10;\sqrt{\dfrac{m}{(k_{1}+k_{2})}}$&#10;$T_{1}=2\pi&#10;\sqrt{\dfrac{m}{k_{1}}}$&#10;$\Rightarrow&#10;k_{1}=(2\pi )^{2}\dfrac{m}{T_{1}^{2}}$&#10;$k_{2}=(2\pi&#10;)^{2}\dfrac{m}{T_{2}^{2}}$&#10;$k_{1}+k_{2}=(2\pi&#10;)^{2}(\dfrac{m}{T_{1}^{2}}+\dfrac{m}{T_{2}^{2}})$&#10;$\therefore&#10;\dfrac{T_{1}T_{2}}{\sqrt{T_{1}+T_{2}}}=T$&#10;&#10;&#10;&#10;

Question 5

An object is attached to the bottom of a light vertical spring and set vibrating. The maximum speed of the object is $15 cm/sec$ and the period is $628$ milliseconds. The amplitude of the motion in centimeters is:

Accepted Answer

&#10;&#10;&#10;&#9;&#10;&#9;&#10;&#9;&#10;&#9;&#10;&#10;&#10;$v_{max}=15\ cm /s&#10;; T=0.628\ s=\dfrac{2\pi}{10}s$$\omega =\dfrac{2\pi }{T}=10\&#10;rad/s$$v_{max}=A\omega $$A=1.5\ cm$&#10;&#10;

Question 6

Assertion (A): The phase difference between displacement and velocity in SHM is

9 0^{\circ}

Reason (R): The displacement is represented by y=A sin

ω

t and Velocity by V=A

ω

cos

ω

t.

Accepted Answer

At mean displacement is Zero but velocity is maximum  At extreme position displacement is maximum but velocity is Zero.Thus displacement can be represented by $y$= $A \ sin \ wt$$V=\dfrac{dy}{dt}$$=Awcoswt= Aw sin (wt+\dfrac{\pi}{2})$Thus phase difference is $\dfrac{\pi}{2}$

Question 7

A man of mass 60 kg, standing on a platform, is executing SHM in a vertical plane. The displacement from mean position is $y = 0.5 sin (2\pi t)$. The minimum value of frequency ($f$) for which the man will feel weightlessness at the highest point is :

Accepted Answer

$w=2\pif$$A=0.5$$a=-w^{2}A$For weightlessness we have $a=g$Thus$g=-w^{2}A$$w^{2}=\dfrac{g}{0.5}=2g$$4\pi^{2}f^{2}=2g$$f^{2}=\dfrac{g}{2\pi^{2}};f=\dfrac{\sqrt{2g}}{2\pi}$

Question 8

If the displacement $x$ and velocity $v$ of a particle executing S.H.M are related through the expression $ 4v^{2}=25-x^{2} $, then its maximum displacement in meters is

Accepted Answer

$v=\omega\sqrt{A^{2}-x^{2}}$$4v^{2}=5^{2} -x^{2} \implies v=\frac{1}{2} \sqrt{(5^{2}-x^{2})}$Comparing the equation with the standard equation we get $A = $5m

Question 9

The displacement equations of two simple harmonic oscillators are given by $ x_{1} =A_{1}\cos\omega t$; $ x_{2}= A_{2} \sin\left(\omega t+\frac{\pi }{6} \right )$. The  phase difference between them is:

Accepted Answer

$X_{1}=A_{1} cos\omega t =A_{1} sin (\omega t+^{\pi}/_{2})$$X_{2}=A_{2} sin (\omega t+^{\pi}/_{6})$so, phase difference $=\dfrac{\pi}{2}-\dfrac{\pi}{6}=\dfrac{\pi}{3}=60^{0}$

Question 10

The amplitude of oscillation of a particle is 0.05 m.If its period is 1.57s. Then the velocity at the mean position is

Accepted Answer

$A=0.05m$$T=\dfrac{\pi }{2}sec ; w=4\ { rad}/{sec}$$v_{max}=wA =0.2 \ {m}/{s}$

Oscillations

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The amplitude of a particle performing S.H.M. is $A$ . The displacement at which its velocity will be half of the maximum velocity is

The equation of the displacement of two particles making SHM are represented by
$y_{1}$ = a sin $(ω t + ϕ)$ & $y_{2}$ = a cos $(ω t)$ .
The phase difference of the velocities of the two particles is :

Acceleration-displacement graph of a particle executing SHM is as shown in the figure. The time period of oscillation is (in sec)

$T_{1}$ , $T_{2}$ are time periods of oscillation of a block individually suspended to spring of force constants $K_{1}$ , $K_{2}$ respectively. If same block is suspended to parallel combination of same two springs, its time period is

An object is attached to the bottom of a light vertical spring and set vibrating. The maximum speed of the object is $15 c m / s e c$ and the period is $628$ milliseconds. The amplitude of the motion in centimeters is:

Assertion (A): The phase difference between displacement and velocity in SHM is $9 0^{\circ}$
Reason (R): The displacement is represented by y=A sin $ω$ t and Velocity by V=A $ω$ cos $ω$ t.

A man of mass 60 kg, standing on a platform, is executing SHM in a vertical plane. The displacement from mean position is $y = 0.5 s i n (2 π t)$ . The minimum value of frequency ( $f$ ) for which the man will feel weightlessness at the highest point is :

If the displacement $x$ and velocity $v$ of a particle executing S.H.M are related through the expression $4 v^{2} = 25 - x^{2}$ , then its maximum displacement in meters is

The displacement equations of two simple harmonic oscillators are given by $x_{1} = A_{1} cos ω t$ ; $x_{2} = A_{2} sin (ω t + \frac{π}{6})$ . The phase difference between them is:

The amplitude of oscillation of a particle is 0.05 m.If its period is 1.57s. Then the velocity at the mean position is

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The amplitude of a particle performing S.H.M. is A. The displacement at which its velocity will be half of the maximum velocity is

The equation of the displacement of two particles making SHM are represented by y1​ = a sin (ωt+ϕ) & y2​ = a cos (ωt). The phase difference of the velocities of the two particles is :

Acceleration-displacement graph of a particle executing SHM is as shown in the figure. The time period of oscillation is (in sec)

T1​, T2​ are time periods of oscillation of a block individually suspended to spring of force constants K1​ ,K2​ respectively. If same block is suspended to parallel combination of same two springs, its time period is

An object is attached to the bottom of a light vertical spring and set vibrating. The maximum speed of the object is 15cm/sec and the period is 628 milliseconds. The amplitude of the motion in centimeters is:

Assertion (A): The phase difference between displacement and velocity in SHM is 90∘ Reason (R): The displacement is represented by y=A sinωt and Velocity by V=Aωcosωt.

A man of mass 60 kg, standing on a platform, is executing SHM in a vertical plane. The displacement from mean position is y=0.5sin(2πt). The minimum value of frequency (f) for which the man will feel weightlessness at the highest point is :

If the displacement x and velocity v of a particle executing S.H.M are related through the expression 4v2=25−x2, then its maximum displacement in meters is

The displacement equations of two simple harmonic oscillators are given by x1​=A1​cosωt; x2​=A2​sin(ωt+6π​). The phase difference between them is:

The amplitude of oscillation of a particle is 0.05 m.If its period is 1.57s. Then the velocity at the mean position is

Physics
5215 Views

The amplitude of a particle performing S.H.M. is $A$ . The displacement at which its velocity will be half of the maximum velocity is

The equation of the displacement of two particles making SHM are represented by
$y_{1}$ = a sin $(ω t + ϕ)$ & $y_{2}$ = a cos $(ω t)$ .
The phase difference of the velocities of the two particles is :

$T_{1}$ , $T_{2}$ are time periods of oscillation of a block individually suspended to spring of force constants $K_{1}$ , $K_{2}$ respectively. If same block is suspended to parallel combination of same two springs, its time period is

An object is attached to the bottom of a light vertical spring and set vibrating. The maximum speed of the object is $15 c m / s e c$ and the period is $628$ milliseconds. The amplitude of the motion in centimeters is:

Assertion (A): The phase difference between displacement and velocity in SHM is $9 0^{\circ}$
Reason (R): The displacement is represented by y=A sin $ω$ t and Velocity by V=A $ω$ cos $ω$ t.

A man of mass 60 kg, standing on a platform, is executing SHM in a vertical plane. The displacement from mean position is $y = 0.5 s i n (2 π t)$ . The minimum value of frequency ( $f$ ) for which the man will feel weightlessness at the highest point is :

If the displacement $x$ and velocity $v$ of a particle executing S.H.M are related through the expression $4 v^{2} = 25 - x^{2}$ , then its maximum displacement in meters is

The displacement equations of two simple harmonic oscillators are given by $x_{1} = A_{1} cos ω t$ ; $x_{2} = A_{2} sin (ω t + \frac{π}{6})$ . The phase difference between them is: