A spring pendulum is suspended from the top of a car. If the car accelerates on a horizontal road, the frequency of oscillation will

be zero

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remain same

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increase

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decrease

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Solution

The correct option is C increase
When the car moves with acceleration a, net effective acceleration of pendulum = $\sqrt{a^{2} + g^{2}}$ if $g = 10 m / s^{2}$
$T = 2 π \sqrt{\frac{l}{a_{e f f}}} ⟹ T \propto \sqrt{\frac{1}{a_{e f f}}} ⟹ f \propto \sqrt{a_{e f f}}$
Thus, the frequency of the oscillation will increase. Hence answer is C

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A spring pendulum is suspended from the top of a car. If the car accelerates on a horizontal road, the frequency of oscillation will

The correct option is C increaseWhen the car moves with acceleration a, net effective acceleration of pendulum = √a2+g2 if g=10m/s2T=2π√laeff⟹T∝√1aeff⟹f∝√aeff Thus, the frequency of the oscillation will increase. Hence answer is C