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M3 - Edexcel - SHM

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    From Edexcel Modular Mathematics Book (Keith Pledger) - Page 62, Qn 9.

    Essentially, mass = 0.4 attached to ends of two identical springs, nat length 1.2m, modulus of elastcitiy =12. Other ends attached to points A & B, a distance of 4m apart, and the mass moved so that AC = 1.4m and CB = 2.6.

    a) Show SHM
    b) Find Max KE

    a) Did the first part, but canot check answer (not in back), I have w^2 as 50.

    I use Ta = 12(1.6-x)/1.2 for one and Tb = 12(1.6+x)/1.2, take one from the other and make resultant F = ma.

    b) I did this part, bu saying max velocity is when displacement is zero, and use v^2=w^2a^2, with a as 0.6 and use KE = 0.5mv^2 -> I get an answer of 6J, in the back it's 3.6J

    Where am I going wrong
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    (Original post by Charries)
    From Edexcel Modular Mathematics Book (Keith Pledger) - Page 62, Qn 9.

    Essentially, mass = 0.4 attached to ends of two identical springs, nat length 1.2m, modulus of elastcitiy =12. Other ends attached to points A & B, a distance of 4m apart, and the mass moved so that AC = 1.4m and CB = 2.6.

    a) Show SHM
    b) Find Max KE

    a) Did the first part, but canot check answer (not in back), I have w^2 as 50.

    I use Ta = 12(1.6-x)/1.2 for one and Tb = 12(1.6+x)/1.2, take one from the other and make resultant F = ma.

    b) I did this part, bu saying max velocity is when displacement is zero, and use v^2=w^2a^2, with a as 0.6 and use KE = 0.5mv^2 -> I get an answer of 6J, in the back it's 3.6J

    Where am I going wrong
     w^2 = 50 \ \ \ \ a=0.6

     v^2=w^2a^2=50 \times 0.6^2

     KE=\frac12mv^2= \frac12 \times 0.4 \times 50 \times 0.6^2 =3.6J

    You have probably made a calculation mistake.
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    OK - thank you.

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