A student measures the maximum speed of a block undergoing simple harmonic oscillations of amplitude A on the end of an ideal spring. If the block is replaced by one with twice the mass but the amplitude of its oscillations remains the same, then the maximum speed of the block will a. decrease by a factor of 2^1/2 b. increase by a factor of 2 c. decrease by a factor of 2 d. decrease by a factor of 4

The maximum speed of a block undergoing simple harmonic oscillations is given by the formula:<br /><br />v_max = ωA<br /><br />where ω is the angular frequency and A is the amplitude of the oscillations.<br /><br />The angular frequency ω is related to the mass m and the spring constant k by the formula:<br /><br />ω = √(k/m)<br /><br />If the block is replaced by one with twice the mass, the new angular frequency ω' is:<br /><br />ω' = √(k/2m) = ω/√2<br /><br />Since the amplitude of the oscillations remains the same, the new maximum speed v_max' is:<br /><br />v_max' = ω' * A = (ω/√2) * A = v_max/√2<br /><br />Therefore, the maximum speed of the block will decrease by a factor of √2.<br /><br />The correct answer is a. decrease by a factor of √2.