A simple pendulum has a period of 4 seconds on earth surface. The length of the

Question

A simple pendulum has a period of 4 seconds on earth surface. The length of the pendulum is: a. 24.95 m b. 6.24 m c. 9.73 m d. 3.97 m e. None of the above A particle moves in simple harmonic motion according to x = 2cos(50f), where x displacement from equilibrium point in meters and t is in seconds. Its maximum speed in m/s is: a. 100 b. 100 sin (50f) c. 100 cos(50f) d. 200 e. none of these A body of mass 10 kg attached with the spring of such stiffness that exerts force 9 N stretches the spring 0.05 m. If the mass is released, what its time period of oscillation? a. 7.23 s b. 10 s c. 5.35 s d. 1.48 s e. none of these

Explanation / Answer

1)

T = 2pi*sqrt(L/g)

L = T^2 g / (4pi^2)

L = (4^2 x 9.81 m/s^2) / (4pi^2)

L = 3.97 m

Ans (d) - 3.97 m

2)

According to the laws of simple harmonic motion,

v(max) = A* = 2*50 = 100 m/s

Ans(a) = 100

3)

Not sure

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