The amplitude of a particle executing S.H.M. with frequency of 60 Hz is 0.01 m. The maximum value of the acceleration of the particle is

[DPMT 1998; CBSE PMT 1999; AFMC 2001;

Pb. PMT 2001; Pb. PET 2001, 02; CPMT 1993, 95, 04;

RPMT 2005; MP PMT 2005]

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Solution

The correct option is A

Maximum acceleration = a $ω^{2}$ = a $\times 4 π^{2} n^{2}$

= $0.01 \times 4 \times (π)^{2} \times (60)^{2} = 144 π^{2} m / s e c^{2}$

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The amplitude of a particle executing S.H.M. with frequency of 60 Hz is 0.01 m. The maximum value of the acceleration of the particle is [DPMT 1998; CBSE PMT 1999; AFMC 2001; Pb. PMT 2001; Pb. PET 2001, 02; CPMT 1993, 95, 04; RPMT 2005; MP PMT 2005]

The correct option is A Maximum acceleration = a ω2 = a ×4π2n2 = 0.01×4×(π)2×(60)2=144π2m/sec2

The amplitude of a particle executing S.H.M. with frequency of 60 Hz is 0.01 m. The maximum value of the acceleration of the particle is

[DPMT 1998; CBSE PMT 1999; AFMC 2001;

Pb. PMT 2001; Pb. PET 2001, 02; CPMT 1993, 95, 04;

RPMT 2005; MP PMT 2005]

The correct option is A

Maximum acceleration = a $ω^{2}$ = a $\times 4 π^{2} n^{2}$

= $0.01 \times 4 \times (π)^{2} \times (60)^{2} = 144 π^{2} m / s e c^{2}$