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PHXI14:OSCILLATIONS

364176 Two simple harmonic motions are represented by $y_{1} = 5 [\sin 2 π t + \sqrt{3} \cos 2 π t]$ and $y_{2} = 5 \sin (2 π t + \frac{π}{4}) .$ The ratio of their amplitudes is

1

1 : 1

2

2 : 1

3

3 : 1

4

\sqrt{3} : 1

Explanation:

$y_{1} = 5 [\sin 2 π t + \sqrt{3} \cos 2 π t]$
$\begin{aligned} = 10 [\frac{1}{2} \sin 2 π t + \frac{\sqrt{3}}{2} \cos 2 π t] \\ = 10 [\cos \frac{π}{3} \sin 2 π t + \cos \frac{π}{3} \cos 2 π t] \\ = 10 [\sin (2 π t + \frac{π}{3})] \end{aligned}$
$∴ A_{1} = 10$
$y_{2} = 5 \sin (2 π t + \frac{π}{4})$
$∴ A_{2} = 5$ , Hence $\frac{A_{1}}{A_{2}} = \frac{2}{1}$

KCET - 2010

PHXI14:OSCILLATIONS

364177 Statement A :
If the amplitude of SHM of a spring mass system is increased, then time period of SHM will remain constant.
Statement B :
If amplitude is increased, body will have to travel more distance to complete one oscillation.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

Time period of oscillation of spring mass system is
$T = 2 π \sqrt{\frac{m}{k}}$
which is independent of the amplitude.
Thus, if the amplitude of the system is increased, then $T$ will remain same. So option (1) is correct.

PHXI14:OSCILLATIONS

364178 A particle performs SHM with amplitude $25 c m$ and period $3 s$ . The minimum time required for it to move between two points $12.5 c m$ on either side of the mean position is

1

0.6 s

2

0.5 s

3

0.4 s

4

0.2 s

Explanation:

Here, amplitude of particle, $A = 25 c m$ and time period, $T = 3 s$
If the particle at $t = 0$ is at mean position, its displacement equation will be
$x = A \sin ω t$
i.e., $x = 25 \sin \frac{2 π t}{3} [∵ ω = 2 π v = \frac{2 π}{T} = \frac{2 π}{3}]$
If it takes time $t_{1}$ to move a distance $x = 12.5 c m$ to one side of its mean position, then
$12.5 = 25 \sin \frac{2 π t_{1}}{3}$
or $\frac{1}{2} = \sin \frac{2 π t_{1}}{3}$ or $\sin \frac{π}{6} = \sin \frac{2 π t_{1}}{3}$
$∴ \frac{π}{6} = \frac{2 π t_{1}}{3} \Rightarrow t_{1} = \frac{1}{4} s$
The same will be the time to move $12.5 c m$ to the other side of its mean position, therefore, total time
$t = t_{1} + t_{2} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} = 0.5 s$

PHXI14:OSCILLATIONS

364179 If $x = 5 \sin (π t + \frac{π}{3}) m$ represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are

1

5 c m, 2 s

2

5 m, 2 s

3

5 c m, 1 s

4

5 m, 1 s

Explanation:

$x = 5 \sin (π t + \frac{π}{3}) m$
comparing the given equation with standard one
$x = A \sin (ω t + ϕ)$
Amplitude $= 5 m$
$ω = π = \frac{2 π}{T}$
$T = \frac{2 π}{π} = 2 s$

NEET - 2024

PHXI14:OSCILLATIONS

364176 Two simple harmonic motions are represented by $y_{1} = 5 [\sin 2 π t + \sqrt{3} \cos 2 π t]$ and $y_{2} = 5 \sin (2 π t + \frac{π}{4}) .$ The ratio of their amplitudes is

1

1 : 1

2

2 : 1

3

3 : 1

4

\sqrt{3} : 1

Explanation:

$y_{1} = 5 [\sin 2 π t + \sqrt{3} \cos 2 π t]$
$\begin{aligned} = 10 [\frac{1}{2} \sin 2 π t + \frac{\sqrt{3}}{2} \cos 2 π t] \\ = 10 [\cos \frac{π}{3} \sin 2 π t + \cos \frac{π}{3} \cos 2 π t] \\ = 10 [\sin (2 π t + \frac{π}{3})] \end{aligned}$
$∴ A_{1} = 10$
$y_{2} = 5 \sin (2 π t + \frac{π}{4})$
$∴ A_{2} = 5$ , Hence $\frac{A_{1}}{A_{2}} = \frac{2}{1}$

KCET - 2010

PHXI14:OSCILLATIONS

364177 Statement A :
If the amplitude of SHM of a spring mass system is increased, then time period of SHM will remain constant.
Statement B :
If amplitude is increased, body will have to travel more distance to complete one oscillation.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

Time period of oscillation of spring mass system is
$T = 2 π \sqrt{\frac{m}{k}}$
which is independent of the amplitude.
Thus, if the amplitude of the system is increased, then $T$ will remain same. So option (1) is correct.

PHXI14:OSCILLATIONS

364178 A particle performs SHM with amplitude $25 c m$ and period $3 s$ . The minimum time required for it to move between two points $12.5 c m$ on either side of the mean position is

1

0.6 s

2

0.5 s

3

0.4 s

4

0.2 s

Explanation:

Here, amplitude of particle, $A = 25 c m$ and time period, $T = 3 s$
If the particle at $t = 0$ is at mean position, its displacement equation will be
$x = A \sin ω t$
i.e., $x = 25 \sin \frac{2 π t}{3} [∵ ω = 2 π v = \frac{2 π}{T} = \frac{2 π}{3}]$
If it takes time $t_{1}$ to move a distance $x = 12.5 c m$ to one side of its mean position, then
$12.5 = 25 \sin \frac{2 π t_{1}}{3}$
or $\frac{1}{2} = \sin \frac{2 π t_{1}}{3}$ or $\sin \frac{π}{6} = \sin \frac{2 π t_{1}}{3}$
$∴ \frac{π}{6} = \frac{2 π t_{1}}{3} \Rightarrow t_{1} = \frac{1}{4} s$
The same will be the time to move $12.5 c m$ to the other side of its mean position, therefore, total time
$t = t_{1} + t_{2} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} = 0.5 s$

PHXI14:OSCILLATIONS

364179 If $x = 5 \sin (π t + \frac{π}{3}) m$ represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are

1

5 c m, 2 s

2

5 m, 2 s

3

5 c m, 1 s

4

5 m, 1 s

Explanation:

$x = 5 \sin (π t + \frac{π}{3}) m$
comparing the given equation with standard one
$x = A \sin (ω t + ϕ)$
Amplitude $= 5 m$
$ω = π = \frac{2 π}{T}$
$T = \frac{2 π}{π} = 2 s$

NEET - 2024

PHXI14:OSCILLATIONS

364176 Two simple harmonic motions are represented by $y_{1} = 5 [\sin 2 π t + \sqrt{3} \cos 2 π t]$ and $y_{2} = 5 \sin (2 π t + \frac{π}{4}) .$ The ratio of their amplitudes is

1

1 : 1

2

2 : 1

3

3 : 1

4

\sqrt{3} : 1

Explanation:

$y_{1} = 5 [\sin 2 π t + \sqrt{3} \cos 2 π t]$
$\begin{aligned} = 10 [\frac{1}{2} \sin 2 π t + \frac{\sqrt{3}}{2} \cos 2 π t] \\ = 10 [\cos \frac{π}{3} \sin 2 π t + \cos \frac{π}{3} \cos 2 π t] \\ = 10 [\sin (2 π t + \frac{π}{3})] \end{aligned}$
$∴ A_{1} = 10$
$y_{2} = 5 \sin (2 π t + \frac{π}{4})$
$∴ A_{2} = 5$ , Hence $\frac{A_{1}}{A_{2}} = \frac{2}{1}$

KCET - 2010

PHXI14:OSCILLATIONS

364177 Statement A :
If the amplitude of SHM of a spring mass system is increased, then time period of SHM will remain constant.
Statement B :
If amplitude is increased, body will have to travel more distance to complete one oscillation.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

Time period of oscillation of spring mass system is
$T = 2 π \sqrt{\frac{m}{k}}$
which is independent of the amplitude.
Thus, if the amplitude of the system is increased, then $T$ will remain same. So option (1) is correct.

PHXI14:OSCILLATIONS

364178 A particle performs SHM with amplitude $25 c m$ and period $3 s$ . The minimum time required for it to move between two points $12.5 c m$ on either side of the mean position is

1

0.6 s

2

0.5 s

3

0.4 s

4

0.2 s

Explanation:

Here, amplitude of particle, $A = 25 c m$ and time period, $T = 3 s$
If the particle at $t = 0$ is at mean position, its displacement equation will be
$x = A \sin ω t$
i.e., $x = 25 \sin \frac{2 π t}{3} [∵ ω = 2 π v = \frac{2 π}{T} = \frac{2 π}{3}]$
If it takes time $t_{1}$ to move a distance $x = 12.5 c m$ to one side of its mean position, then
$12.5 = 25 \sin \frac{2 π t_{1}}{3}$
or $\frac{1}{2} = \sin \frac{2 π t_{1}}{3}$ or $\sin \frac{π}{6} = \sin \frac{2 π t_{1}}{3}$
$∴ \frac{π}{6} = \frac{2 π t_{1}}{3} \Rightarrow t_{1} = \frac{1}{4} s$
The same will be the time to move $12.5 c m$ to the other side of its mean position, therefore, total time
$t = t_{1} + t_{2} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} = 0.5 s$

PHXI14:OSCILLATIONS

364179 If $x = 5 \sin (π t + \frac{π}{3}) m$ represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are

1

5 c m, 2 s

2

5 m, 2 s

3

5 c m, 1 s

4

5 m, 1 s

Explanation:

$x = 5 \sin (π t + \frac{π}{3}) m$
comparing the given equation with standard one
$x = A \sin (ω t + ϕ)$
Amplitude $= 5 m$
$ω = π = \frac{2 π}{T}$
$T = \frac{2 π}{π} = 2 s$

NEET - 2024

PHXI14:OSCILLATIONS

364176 Two simple harmonic motions are represented by $y_{1} = 5 [\sin 2 π t + \sqrt{3} \cos 2 π t]$ and $y_{2} = 5 \sin (2 π t + \frac{π}{4}) .$ The ratio of their amplitudes is

1

1 : 1

2

2 : 1

3

3 : 1

4

\sqrt{3} : 1

Explanation:

$y_{1} = 5 [\sin 2 π t + \sqrt{3} \cos 2 π t]$
$\begin{aligned} = 10 [\frac{1}{2} \sin 2 π t + \frac{\sqrt{3}}{2} \cos 2 π t] \\ = 10 [\cos \frac{π}{3} \sin 2 π t + \cos \frac{π}{3} \cos 2 π t] \\ = 10 [\sin (2 π t + \frac{π}{3})] \end{aligned}$
$∴ A_{1} = 10$
$y_{2} = 5 \sin (2 π t + \frac{π}{4})$
$∴ A_{2} = 5$ , Hence $\frac{A_{1}}{A_{2}} = \frac{2}{1}$

KCET - 2010

PHXI14:OSCILLATIONS

364177 Statement A :
If the amplitude of SHM of a spring mass system is increased, then time period of SHM will remain constant.
Statement B :
If amplitude is increased, body will have to travel more distance to complete one oscillation.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

Time period of oscillation of spring mass system is
$T = 2 π \sqrt{\frac{m}{k}}$
which is independent of the amplitude.
Thus, if the amplitude of the system is increased, then $T$ will remain same. So option (1) is correct.

PHXI14:OSCILLATIONS

364178 A particle performs SHM with amplitude $25 c m$ and period $3 s$ . The minimum time required for it to move between two points $12.5 c m$ on either side of the mean position is

1

0.6 s

2

0.5 s

3

0.4 s

4

0.2 s

Explanation:

Here, amplitude of particle, $A = 25 c m$ and time period, $T = 3 s$
If the particle at $t = 0$ is at mean position, its displacement equation will be
$x = A \sin ω t$
i.e., $x = 25 \sin \frac{2 π t}{3} [∵ ω = 2 π v = \frac{2 π}{T} = \frac{2 π}{3}]$
If it takes time $t_{1}$ to move a distance $x = 12.5 c m$ to one side of its mean position, then
$12.5 = 25 \sin \frac{2 π t_{1}}{3}$
or $\frac{1}{2} = \sin \frac{2 π t_{1}}{3}$ or $\sin \frac{π}{6} = \sin \frac{2 π t_{1}}{3}$
$∴ \frac{π}{6} = \frac{2 π t_{1}}{3} \Rightarrow t_{1} = \frac{1}{4} s$
The same will be the time to move $12.5 c m$ to the other side of its mean position, therefore, total time
$t = t_{1} + t_{2} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} = 0.5 s$

PHXI14:OSCILLATIONS

364179 If $x = 5 \sin (π t + \frac{π}{3}) m$ represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are

1

5 c m, 2 s

2

5 m, 2 s

3

5 c m, 1 s

4

5 m, 1 s

Explanation:

$x = 5 \sin (π t + \frac{π}{3}) m$
comparing the given equation with standard one
$x = A \sin (ω t + ϕ)$
Amplitude $= 5 m$
$ω = π = \frac{2 π}{T}$
$T = \frac{2 π}{π} = 2 s$

NEET - 2024

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