JEE MAIN - Physics (2016 - 10th April Morning Slot - No. 14)
In an engine the piston undergoes vertical simple harmonic motion with amplitude 7 cm. A washer rests on top of the piston and moves with it. The motor speed is slowly increased. The frequency of the piston at which the washer no longer stays in contact with the piston, is close to :
0.1 Hz
1.2 Hz
0.7 Hz
1.9 Hz
Selitys
Here,
Amplitude, A = 7 cm = 0.07 m
When washer is no longer stays in contact with the piston, then the normal force on the washer is = 0
$$ \therefore $$ Maximum acceleration of the washer,
amax = $$\omega $$2A = g
$$ \Rightarrow $$ $$\omega $$ = $$\sqrt {{g \over A}} $$ = $$\sqrt {{{10} \over {0.07}}} $$ = $$\sqrt {{{1000} \over 7}} $$
$$ \therefore $$ Frequency of the piston,
f = $${\omega \over {2\pi }}$$ = $${1 \over {2\pi }}\sqrt {{{1000} \over 7}} $$ = 1.9 Hz
Amplitude, A = 7 cm = 0.07 m
When washer is no longer stays in contact with the piston, then the normal force on the washer is = 0
$$ \therefore $$ Maximum acceleration of the washer,
amax = $$\omega $$2A = g
$$ \Rightarrow $$ $$\omega $$ = $$\sqrt {{g \over A}} $$ = $$\sqrt {{{10} \over {0.07}}} $$ = $$\sqrt {{{1000} \over 7}} $$
$$ \therefore $$ Frequency of the piston,
f = $${\omega \over {2\pi }}$$ = $${1 \over {2\pi }}\sqrt {{{1000} \over 7}} $$ = 1.9 Hz