A small body of mass 0.10 kg is executing S.H.M. of amplitude 1.0 m and period 0.20 sec. The maximum force acting on it is

98.596 N

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985.96 N

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100.2 N

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76.23 N

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Solution

The correct option is A
98.596 N

Maximum acceleration

$A_{m a x} = a ω^{2} = \frac{a \times 4 π^{2}}{T^{2}} = \frac{1 \times 4 \times (3.14)^{2}}{0.2 \times 0.2}$

$F_{m a x} = m \times A_{m a x} = \frac{0.1 \times 4 \times (3.14)^{2}}{0.2 \times 0.2}$ = 98.596 N

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A small body of mass 0.10 kg is executing S.H.M. of amplitude 1.0 m and period 0.20 sec. The maximum force acting on it is

The correct option is A 98.596 N Maximum acceleration Amax=aω2=a×4π2T2=1×4×(3.14)20.2×0.2 Fmax=m×Amax=0.1×4×(3.14)20.2×0.2 = 98.596 N

The correct option is A
98.596 N

Maximum acceleration

$A_{m a x} = a ω^{2} = \frac{a \times 4 π^{2}}{T^{2}} = \frac{1 \times 4 \times (3.14)^{2}}{0.2 \times 0.2}$

$F_{m a x} = m \times A_{m a x} = \frac{0.1 \times 4 \times (3.14)^{2}}{0.2 \times 0.2}$ = 98.596 N