JEE Main 2015 (Offline) | Simple Harmonic Motion Question 134 | Physics

The period of oscillation of a simple pendulum is $$T = 2\pi \sqrt {{L \over g}} $$. Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using wrist watch of 1 s resolution. The accuracy in the determination of g is:

1 %

5 %

2 %

3 %

JEE Main 2015 (Offline)

MCQ (Single Correct Answer)

-1

For a simple pendulum, a graph is plotted between its kinetic energy $$(KE)$$ and potential energy $$(PE)$$ against its displacement $$d.$$ Which one of the following represents these correctly?
$$(graphs$$ $$are$$ $$schematic$$ $$and$$ $$not$$ $$drawn$$ $$to$$ $$scale)$$

JEE Main 2015 (Offline) Physics - Simple Harmonic Motion Question 134 English Option 1

JEE Main 2015 (Offline) Physics - Simple Harmonic Motion Question 134 English Option 2

JEE Main 2015 (Offline)

MCQ (Single Correct Answer)

-1

A pendulum made of a uniform wire of cross sectional area $$A$$ has time period $$T.$$ When an additional mass $$M$$ is added to its bob, the time period changes to $${T_{M.}}$$ If the Young's modulus of the material of the wire is $$Y$$ then $${1 \over Y}$$ is equal to :
($$g=$$ $$gravitational$$ $$acceleration$$)

$$\left[ {1 - {{\left( {{{{T_M}} \over T}} \right)}^2}} \right]{A \over {Mg}}$$

$$\left[ {1 - {{\left( {{T \over {{T_M}}}} \right)}^2}} \right]{A \over {Mg}}$$

$$\left[ {{{\left( {{{{T_M}} \over T}} \right)}^2} - 1} \right]{A \over {Mg}}$$

$$\left[ {{{\left( {{{{T_M}} \over T}} \right)}^2} - 1} \right]{{Mg} \over A}$$

JEE Main 2014 (Offline)

MCQ (Single Correct Answer)

-1

A particle moves with simple harmonic motion in a straight line. In first $$\tau s,$$ after starting from rest it travels a distance $$a,$$ and in next $$\tau s$$ it travels $$2a,$$ in same direction, then:

amplitude of motion is $$3a$$

time period of oscillations is $$8\tau $$

amplitude of motion is $$4a$$

time period of oscillations is $$6\tau $$

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