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PHXI14:OSCILLATIONS

364240 Statement A :
Simple harmonic motion is the projection of uniform circular motion on the diameter of a circle.
Statement B :
Simple harmonic motion is a uniform motion.

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.

PHXI14:OSCILLATIONS

364241 The circular motion of a particle with constant speed is

1 periodic but not SHM

2 SHM but not periodic

3 periodic and also SHM

4 neither periodic nor SHM.

KCET - 2012

PHXI14:OSCILLATIONS

364242 Which of the following four statements is false?

1 A body can have zero velocity and still be accelerated

2 A body can have a constant velocity and still have a varying speed

3 A body can have a constant speed and still have a varying velocity

4 The direction of the velocity of a body can change when its speed is constant

PHXI14:OSCILLATIONS

364243 In S.H.M maximum acceleration is at

1 Equilibrium

2 Extreme position

3 Acceleration is constant

4 None of these

Explanation:

$A_{max} = ω^{2} a$ .

PHXI14:OSCILLATIONS

364244 A particle is performing a linear simple harmonic motion of amplitude ' $A$ '. When it is midway between its mean and extreme position, the magnitude of its velocity and acceleration are equal. What is the periodic time of the motion?

1

\frac{2 π}{\sqrt{3}} s

2

\frac{\sqrt{3}}{2 π} s

3

2 π \sqrt{3} s

4

\frac{1}{2 π \sqrt{3}} s

Explanation:

In linear simple harmonic motion, the velocity of particle is given by
$v = ω \sqrt{A^{2} - x^{2}} (1)$
where, $ω =$ angular frequency
$A =$ maximum displacement of amplitude and
$x =$ displacement from mean position
The acceleration of a particle in simple harmonic motion(SHM), is given by
$a = ω^{2} x (2)$
Here, $x = \frac{A}{2}$
Alos, $v = a$ (given)
$ω \sqrt{(A^{2} - x^{2})} = ω^{2} x$ from eq.(1) and (2), we get
$\Rightarrow \sqrt{(A^{2} - \frac{A^{2}}{4})} = ω \frac{A}{2} \Rightarrow \frac{\sqrt{3} A}{2} = ω \times \frac{A}{2}$
$\Rightarrow \frac{2 π}{T} = \sqrt{3} [∵ ω = \frac{2 π}{T}]$
$\Rightarrow T = \frac{2 π}{\sqrt{3}} s$

MHTCET - 2019

PHXI14:OSCILLATIONS

364240 Statement A :
Simple harmonic motion is the projection of uniform circular motion on the diameter of a circle.
Statement B :
Simple harmonic motion is a uniform motion.

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.

PHXI14:OSCILLATIONS

364241 The circular motion of a particle with constant speed is

1 periodic but not SHM

2 SHM but not periodic

3 periodic and also SHM

4 neither periodic nor SHM.

KCET - 2012

PHXI14:OSCILLATIONS

364242 Which of the following four statements is false?

1 A body can have zero velocity and still be accelerated

2 A body can have a constant velocity and still have a varying speed

3 A body can have a constant speed and still have a varying velocity

4 The direction of the velocity of a body can change when its speed is constant

PHXI14:OSCILLATIONS

364243 In S.H.M maximum acceleration is at

1 Equilibrium

2 Extreme position

3 Acceleration is constant

4 None of these

Explanation:

$A_{max} = ω^{2} a$ .

PHXI14:OSCILLATIONS

364244 A particle is performing a linear simple harmonic motion of amplitude ' $A$ '. When it is midway between its mean and extreme position, the magnitude of its velocity and acceleration are equal. What is the periodic time of the motion?

1

\frac{2 π}{\sqrt{3}} s

2

\frac{\sqrt{3}}{2 π} s

3

2 π \sqrt{3} s

4

\frac{1}{2 π \sqrt{3}} s

Explanation:

In linear simple harmonic motion, the velocity of particle is given by
$v = ω \sqrt{A^{2} - x^{2}} (1)$
where, $ω =$ angular frequency
$A =$ maximum displacement of amplitude and
$x =$ displacement from mean position
The acceleration of a particle in simple harmonic motion(SHM), is given by
$a = ω^{2} x (2)$
Here, $x = \frac{A}{2}$
Alos, $v = a$ (given)
$ω \sqrt{(A^{2} - x^{2})} = ω^{2} x$ from eq.(1) and (2), we get
$\Rightarrow \sqrt{(A^{2} - \frac{A^{2}}{4})} = ω \frac{A}{2} \Rightarrow \frac{\sqrt{3} A}{2} = ω \times \frac{A}{2}$
$\Rightarrow \frac{2 π}{T} = \sqrt{3} [∵ ω = \frac{2 π}{T}]$
$\Rightarrow T = \frac{2 π}{\sqrt{3}} s$

MHTCET - 2019

PHXI14:OSCILLATIONS

364240 Statement A :
Simple harmonic motion is the projection of uniform circular motion on the diameter of a circle.
Statement B :
Simple harmonic motion is a uniform motion.

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.

PHXI14:OSCILLATIONS

364241 The circular motion of a particle with constant speed is

1 periodic but not SHM

2 SHM but not periodic

3 periodic and also SHM

4 neither periodic nor SHM.

KCET - 2012

PHXI14:OSCILLATIONS

364242 Which of the following four statements is false?

1 A body can have zero velocity and still be accelerated

2 A body can have a constant velocity and still have a varying speed

3 A body can have a constant speed and still have a varying velocity

4 The direction of the velocity of a body can change when its speed is constant

PHXI14:OSCILLATIONS

364243 In S.H.M maximum acceleration is at

1 Equilibrium

2 Extreme position

3 Acceleration is constant

4 None of these

Explanation:

$A_{max} = ω^{2} a$ .

PHXI14:OSCILLATIONS

364244 A particle is performing a linear simple harmonic motion of amplitude ' $A$ '. When it is midway between its mean and extreme position, the magnitude of its velocity and acceleration are equal. What is the periodic time of the motion?

1

\frac{2 π}{\sqrt{3}} s

2

\frac{\sqrt{3}}{2 π} s

3

2 π \sqrt{3} s

4

\frac{1}{2 π \sqrt{3}} s

Explanation:

In linear simple harmonic motion, the velocity of particle is given by
$v = ω \sqrt{A^{2} - x^{2}} (1)$
where, $ω =$ angular frequency
$A =$ maximum displacement of amplitude and
$x =$ displacement from mean position
The acceleration of a particle in simple harmonic motion(SHM), is given by
$a = ω^{2} x (2)$
Here, $x = \frac{A}{2}$
Alos, $v = a$ (given)
$ω \sqrt{(A^{2} - x^{2})} = ω^{2} x$ from eq.(1) and (2), we get
$\Rightarrow \sqrt{(A^{2} - \frac{A^{2}}{4})} = ω \frac{A}{2} \Rightarrow \frac{\sqrt{3} A}{2} = ω \times \frac{A}{2}$
$\Rightarrow \frac{2 π}{T} = \sqrt{3} [∵ ω = \frac{2 π}{T}]$
$\Rightarrow T = \frac{2 π}{\sqrt{3}} s$

MHTCET - 2019

PHXI14:OSCILLATIONS

364240 Statement A :
Simple harmonic motion is the projection of uniform circular motion on the diameter of a circle.
Statement B :
Simple harmonic motion is a uniform motion.

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.

PHXI14:OSCILLATIONS

364241 The circular motion of a particle with constant speed is

1 periodic but not SHM

2 SHM but not periodic

3 periodic and also SHM

4 neither periodic nor SHM.

KCET - 2012

PHXI14:OSCILLATIONS

364242 Which of the following four statements is false?

1 A body can have zero velocity and still be accelerated

2 A body can have a constant velocity and still have a varying speed

3 A body can have a constant speed and still have a varying velocity

4 The direction of the velocity of a body can change when its speed is constant

PHXI14:OSCILLATIONS

364243 In S.H.M maximum acceleration is at

1 Equilibrium

2 Extreme position

3 Acceleration is constant

4 None of these

Explanation:

$A_{max} = ω^{2} a$ .

PHXI14:OSCILLATIONS

364244 A particle is performing a linear simple harmonic motion of amplitude ' $A$ '. When it is midway between its mean and extreme position, the magnitude of its velocity and acceleration are equal. What is the periodic time of the motion?

1

\frac{2 π}{\sqrt{3}} s

2

\frac{\sqrt{3}}{2 π} s

3

2 π \sqrt{3} s

4

\frac{1}{2 π \sqrt{3}} s

Explanation:

In linear simple harmonic motion, the velocity of particle is given by
$v = ω \sqrt{A^{2} - x^{2}} (1)$
where, $ω =$ angular frequency
$A =$ maximum displacement of amplitude and
$x =$ displacement from mean position
The acceleration of a particle in simple harmonic motion(SHM), is given by
$a = ω^{2} x (2)$
Here, $x = \frac{A}{2}$
Alos, $v = a$ (given)
$ω \sqrt{(A^{2} - x^{2})} = ω^{2} x$ from eq.(1) and (2), we get
$\Rightarrow \sqrt{(A^{2} - \frac{A^{2}}{4})} = ω \frac{A}{2} \Rightarrow \frac{\sqrt{3} A}{2} = ω \times \frac{A}{2}$
$\Rightarrow \frac{2 π}{T} = \sqrt{3} [∵ ω = \frac{2 π}{T}]$
$\Rightarrow T = \frac{2 π}{\sqrt{3}} s$

MHTCET - 2019

PHXI14:OSCILLATIONS

364240 Statement A :
Simple harmonic motion is the projection of uniform circular motion on the diameter of a circle.
Statement B :
Simple harmonic motion is a uniform motion.

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.

PHXI14:OSCILLATIONS

364241 The circular motion of a particle with constant speed is

1 periodic but not SHM

2 SHM but not periodic

3 periodic and also SHM

4 neither periodic nor SHM.

KCET - 2012

PHXI14:OSCILLATIONS

364242 Which of the following four statements is false?

1 A body can have zero velocity and still be accelerated

2 A body can have a constant velocity and still have a varying speed

3 A body can have a constant speed and still have a varying velocity

4 The direction of the velocity of a body can change when its speed is constant

PHXI14:OSCILLATIONS

364243 In S.H.M maximum acceleration is at

1 Equilibrium

2 Extreme position

3 Acceleration is constant

4 None of these

Explanation:

$A_{max} = ω^{2} a$ .

PHXI14:OSCILLATIONS

364244 A particle is performing a linear simple harmonic motion of amplitude ' $A$ '. When it is midway between its mean and extreme position, the magnitude of its velocity and acceleration are equal. What is the periodic time of the motion?

1

\frac{2 π}{\sqrt{3}} s

2

\frac{\sqrt{3}}{2 π} s

3

2 π \sqrt{3} s

4

\frac{1}{2 π \sqrt{3}} s

Explanation:

In linear simple harmonic motion, the velocity of particle is given by
$v = ω \sqrt{A^{2} - x^{2}} (1)$
where, $ω =$ angular frequency
$A =$ maximum displacement of amplitude and
$x =$ displacement from mean position
The acceleration of a particle in simple harmonic motion(SHM), is given by
$a = ω^{2} x (2)$
Here, $x = \frac{A}{2}$
Alos, $v = a$ (given)
$ω \sqrt{(A^{2} - x^{2})} = ω^{2} x$ from eq.(1) and (2), we get
$\Rightarrow \sqrt{(A^{2} - \frac{A^{2}}{4})} = ω \frac{A}{2} \Rightarrow \frac{\sqrt{3} A}{2} = ω \times \frac{A}{2}$
$\Rightarrow \frac{2 π}{T} = \sqrt{3} [∵ ω = \frac{2 π}{T}]$
$\Rightarrow T = \frac{2 π}{\sqrt{3}} s$

MHTCET - 2019