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PHXI14:OSCILLATIONS

364279 Average velocity of a particle executing SHM in one complete vibration is :

1

\frac{A ω}{2}

2

A ω

3

\frac{A ω^{2}}{2}

4 Zero

Explanation:

In one complete vibration, displacement is zero. So, average velocity in one complete vibration
$= \frac{D i s p l a c e m e n t}{T i m e i n t e r v a l} = 0$

NEET - 2019

PHXI14:OSCILLATIONS

364280 A $1.00 \times 10^{- 20} k g$ particle is executing with simple harmonic motion with a period of $1.00 \times 10^{- 5} \sec$ and a maximum speed of $1.00 \times 10^{3} m / s$ . The maximum displacement of the particle is:

1

1.59 m m

2

1.00 m

3

10 m

4 None of these

Explanation:

$v_{max} = A ω = A \times \frac{2 π}{T} \Rightarrow A = \frac{v_{max} \times T}{2 π}$
$A = \frac{1.00 \times 10^{3} \times (1 \times 10^{- 5})}{2 π} = 1.59 m m$

PHXI14:OSCILLATIONS

364281 If displacement $x$ and velocity $v$ are related as $4 v^{2} = 25 - x^{2}$ in a SHM then time period of given SHM is (consider SI units)

1

2 π

2

π

3

6 π

4

4 π

Explanation:

Given $4 v^{2} = 25 - x^{2}$
Differentiating with respect to $t$ on both sides,
$8 v a = - 2 x v \Rightarrow a = - \frac{x}{4}$
Motion is simple harmonic:
So, $ω^{2} = \frac{1}{4}$ and $T = 2 π \sqrt{4} = 4 π$

PHXI14:OSCILLATIONS

364282 For a particle executing simple harmonic motion (SHM), at its mean position

1 Velocity is zero and acceleration is maximum

2 Velocity is maximum and acceleration is zero

3 Both velocity and acceleration are maximum

4 Both velocity and acceleration are zero

Explanation:

The relation between velocity, acceleration and displacement is given by
$V = ω \sqrt{A^{2} - y^{2}}$
$a = - ω^{2} y$
At mean position $y = 0$
$∴ V =$ maximum and $a =$ minimum.

KCET - 2024

PHXI14:OSCILLATIONS

364279 Average velocity of a particle executing SHM in one complete vibration is :

1

\frac{A ω}{2}

2

A ω

3

\frac{A ω^{2}}{2}

4 Zero

Explanation:

In one complete vibration, displacement is zero. So, average velocity in one complete vibration
$= \frac{D i s p l a c e m e n t}{T i m e i n t e r v a l} = 0$

NEET - 2019

PHXI14:OSCILLATIONS

364280 A $1.00 \times 10^{- 20} k g$ particle is executing with simple harmonic motion with a period of $1.00 \times 10^{- 5} \sec$ and a maximum speed of $1.00 \times 10^{3} m / s$ . The maximum displacement of the particle is:

1

1.59 m m

2

1.00 m

3

10 m

4 None of these

Explanation:

$v_{max} = A ω = A \times \frac{2 π}{T} \Rightarrow A = \frac{v_{max} \times T}{2 π}$
$A = \frac{1.00 \times 10^{3} \times (1 \times 10^{- 5})}{2 π} = 1.59 m m$

PHXI14:OSCILLATIONS

364281 If displacement $x$ and velocity $v$ are related as $4 v^{2} = 25 - x^{2}$ in a SHM then time period of given SHM is (consider SI units)

1

2 π

2

π

3

6 π

4

4 π

Explanation:

Given $4 v^{2} = 25 - x^{2}$
Differentiating with respect to $t$ on both sides,
$8 v a = - 2 x v \Rightarrow a = - \frac{x}{4}$
Motion is simple harmonic:
So, $ω^{2} = \frac{1}{4}$ and $T = 2 π \sqrt{4} = 4 π$

PHXI14:OSCILLATIONS

364282 For a particle executing simple harmonic motion (SHM), at its mean position

1 Velocity is zero and acceleration is maximum

2 Velocity is maximum and acceleration is zero

3 Both velocity and acceleration are maximum

4 Both velocity and acceleration are zero

Explanation:

The relation between velocity, acceleration and displacement is given by
$V = ω \sqrt{A^{2} - y^{2}}$
$a = - ω^{2} y$
At mean position $y = 0$
$∴ V =$ maximum and $a =$ minimum.

KCET - 2024

PHXI14:OSCILLATIONS

364279 Average velocity of a particle executing SHM in one complete vibration is :

1

\frac{A ω}{2}

2

A ω

3

\frac{A ω^{2}}{2}

4 Zero

Explanation:

In one complete vibration, displacement is zero. So, average velocity in one complete vibration
$= \frac{D i s p l a c e m e n t}{T i m e i n t e r v a l} = 0$

NEET - 2019

PHXI14:OSCILLATIONS

364280 A $1.00 \times 10^{- 20} k g$ particle is executing with simple harmonic motion with a period of $1.00 \times 10^{- 5} \sec$ and a maximum speed of $1.00 \times 10^{3} m / s$ . The maximum displacement of the particle is:

1

1.59 m m

2

1.00 m

3

10 m

4 None of these

Explanation:

$v_{max} = A ω = A \times \frac{2 π}{T} \Rightarrow A = \frac{v_{max} \times T}{2 π}$
$A = \frac{1.00 \times 10^{3} \times (1 \times 10^{- 5})}{2 π} = 1.59 m m$

PHXI14:OSCILLATIONS

364281 If displacement $x$ and velocity $v$ are related as $4 v^{2} = 25 - x^{2}$ in a SHM then time period of given SHM is (consider SI units)

1

2 π

2

π

3

6 π

4

4 π

Explanation:

Given $4 v^{2} = 25 - x^{2}$
Differentiating with respect to $t$ on both sides,
$8 v a = - 2 x v \Rightarrow a = - \frac{x}{4}$
Motion is simple harmonic:
So, $ω^{2} = \frac{1}{4}$ and $T = 2 π \sqrt{4} = 4 π$

PHXI14:OSCILLATIONS

364282 For a particle executing simple harmonic motion (SHM), at its mean position

1 Velocity is zero and acceleration is maximum

2 Velocity is maximum and acceleration is zero

3 Both velocity and acceleration are maximum

4 Both velocity and acceleration are zero

Explanation:

The relation between velocity, acceleration and displacement is given by
$V = ω \sqrt{A^{2} - y^{2}}$
$a = - ω^{2} y$
At mean position $y = 0$
$∴ V =$ maximum and $a =$ minimum.

KCET - 2024

PHXI14:OSCILLATIONS

364279 Average velocity of a particle executing SHM in one complete vibration is :

1

\frac{A ω}{2}

2

A ω

3

\frac{A ω^{2}}{2}

4 Zero

Explanation:

In one complete vibration, displacement is zero. So, average velocity in one complete vibration
$= \frac{D i s p l a c e m e n t}{T i m e i n t e r v a l} = 0$

NEET - 2019

PHXI14:OSCILLATIONS

364280 A $1.00 \times 10^{- 20} k g$ particle is executing with simple harmonic motion with a period of $1.00 \times 10^{- 5} \sec$ and a maximum speed of $1.00 \times 10^{3} m / s$ . The maximum displacement of the particle is:

1

1.59 m m

2

1.00 m

3

10 m

4 None of these

Explanation:

$v_{max} = A ω = A \times \frac{2 π}{T} \Rightarrow A = \frac{v_{max} \times T}{2 π}$
$A = \frac{1.00 \times 10^{3} \times (1 \times 10^{- 5})}{2 π} = 1.59 m m$

PHXI14:OSCILLATIONS

364281 If displacement $x$ and velocity $v$ are related as $4 v^{2} = 25 - x^{2}$ in a SHM then time period of given SHM is (consider SI units)

1

2 π

2

π

3

6 π

4

4 π

Explanation:

Given $4 v^{2} = 25 - x^{2}$
Differentiating with respect to $t$ on both sides,
$8 v a = - 2 x v \Rightarrow a = - \frac{x}{4}$
Motion is simple harmonic:
So, $ω^{2} = \frac{1}{4}$ and $T = 2 π \sqrt{4} = 4 π$

PHXI14:OSCILLATIONS

364282 For a particle executing simple harmonic motion (SHM), at its mean position

1 Velocity is zero and acceleration is maximum

2 Velocity is maximum and acceleration is zero

3 Both velocity and acceleration are maximum

4 Both velocity and acceleration are zero

Explanation:

The relation between velocity, acceleration and displacement is given by
$V = ω \sqrt{A^{2} - y^{2}}$
$a = - ω^{2} y$
At mean position $y = 0$
$∴ V =$ maximum and $a =$ minimum.

KCET - 2024