Oscillations Test

Result

Oscillations Test - 65

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Weekly Quiz Competition

Question 1
1 / -0

The ratio of the total energies of a oscillating particle and its foot of the perpendicular is

A
2:1

B
1:2

C
1:1

D
$$\sqrt{2}$$:1
Question 2
1 / -0

Two $$SHM's$$ with same amplitude and time period, when acting together in perpendicular directions with a phase difference of $$\dfrac{\pi}{2}$$, give rise to

A
Straight motion

B
Elliptical motion

C
Circular motion

D
None of these

Solution
Question 3
1 / -0

A particle is executing $$SHM$$ along a straight line. Its velocities at distance $${x}_{1}$$ and $${x}_{2}$$ from the mean position are $${v}_{1}$$ and $${v}_{2}$$ respectively. Its time period is :

A
$$2\pi \sqrt { \dfrac { { x }_{ 2 }^{ 2 }-{ x }_{ 1 }^{ 2 } }{ { v }_{ 1 }^{ 2 }-{ v }_{ 2 }^{ 2 } } }$$

B
$$2\pi \sqrt { \dfrac { { v }_{ 1 }^{ 2 }+{ v }_{ 2 }^{ 2 } }{ { x }_{ 1 }^{ 2 }-{ x }_{ 2 }^{ 2 } } }$$

C
$$2\pi \sqrt { \dfrac { { v }_{ 1 }^{ 2 }-{ v }_{ 2 }^{ 2 } }{ { x }_{ 1 }^{ 2 }-{ x }_{ 2 }^{ 2 } } }$$

D
$$2\pi \sqrt { \dfrac { { x }_{ 1 }^{ 2 }-{ x }_{ 2 }^{ 2 } }{ { v }_{ 1 }^{ 2 }-{ v }_{ 2 }^{ 2 } } }$$

Solution
Question 4
1 / -0

Find the angular frequency of small oscillation of block m in the arrangement shown. Rod is massless. [Assume gravity to be absent]

A
$$\sqrt{9k_{1} k_{3} + k_{1}k_{2} + 4k_{2}k_{3}/(4k_{3} + k_{2})m}$$

B
$$\sqrt{4k_{1} k_{3} + k_{1}k_{2} + 4k_{2}k_{3}/(4k_{3} + k_{2})m}$$

C
$$\sqrt{8k_{1} k_{3} + k_{1}k_{2} + 4k_{2}k_{3}/(4k_{3} + k_{2})m}$$

D
$$\sqrt{5k_{1} k_{3} + k_{1}k_{2} + 4k_{2}k_{3}/(4k_{3} + k_{2})m}$$

Solution

$$\sqrt{4k_{1} k_{3} + k_{1}k_{2} + 4k_{2}k_{3}/(4k_{3} + k_{2})m}$$
Question 5
1 / -0

A particle is executing SHM along a straight line. Its velocities at distances $$x_1 $$ and $$x_2$$ from the mean position are $$V_1$$ and $$V_2$$ respectively. Its time period is:

A
$$2\pi \sqrt\frac{V_1^2 - V_2^2}{x_1^2 - x_2^2}$$

B
$$2\pi \sqrt\frac{x_1^2 + x_2^2}{V_1^2 + V_2^2}$$

C
$$2\pi \sqrt\frac{x_1^2 - x_2^2}{V_2^2 - V_1^2}$$

D
$$2\pi \sqrt\frac{V_1^2 + V_2^2}{x_1^2 + x_2^2}$$

Solution

$$v=\omega\sqrt{A^2-x^2}$$
$$v^2=\omega^2(A^2-x^2)$$ then,
$$v_1=\omega^2(A^2-x_1^2)$$...(1)
$$v_2=\omega^2(A^2-x_2^2)$$...(2)
Subtracting (1) from (2), we get,
$$v_2^2-v_1^2=\omega^2(x_1^2-x_2^2)$$
$$\omega=\dfrac{\sqrt{v_2^2-v_1^2}}{\sqrt{x_1^2-x_2^2}}$$
$$T=\dfrac{2 \pi}{\omega}=\dfrac{2\pi \sqrt{x_1^2-x_2^2}}{\sqrt{v_2^2-v_1^2}}$$
Question 6
1 / -0

A particle moves on x-axis according to the equation x = $$x_{0} sin^{2} \omega t$$, the motion is simple harmonic

A
with amplitude $$x_{0}$$

B
with amplitude $$2x_{0}$$

C
with time period $$\displaystyle\frac{2\pi}{\omega}$$

D
with time period $$\displaystyle\frac{\pi}{\omega}$$

Solution
Question 7
1 / -0

STATEMENT-1: Time period of simple harmonic motion at moon is more than that on earth because
STATEMENT-2: There is no atmosphere on moon.

A
Statement-1 is True, Statement-2 is True; Statement -2 is a correct explanation for Statement-1.

B
Statement-1 is True, Statement-2 is True; Statement -2 is NOT a correct explanation for Statement-1.

C
Statement -1 is True, Statement-2 is False.

D
Statement -1 is False, Statement-2 is True.

Solution
Question 8
1 / -0

The maximum value attained by the tension in the string of a
swinging pendulum is four times the minimum value it attains. There is no stack
in the string. The angular amplitude of the pendulum is

A
$${90^0}$$

B
$${60^0}$$

C
$${45^0}$$

D
$${30^0}$$

Solution
Question 9
1 / -0

Projection of uniform circular motion on a diameter is:-

A
simple harmonic motion

B
angular simple harmonic motion

C
both a and b

D
None of these.

Solution
Question 10
1 / -0

The string of a pendulum is horizontal initially. The mass of the bob attached is m. Now the string is released. 'R' is the length of the pendulum.
(i) Velocity of the string at the lowest point

A
$$\sqrt{2 gR}$$

B
$$\sqrt{3 gR}$$

C
$$\sqrt{4 gR}$$

D
$$\sqrt{5 gR}$$