IIT-JEE 2009 · Paper 2 · Physics Q57

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PhysicsInteger+4 / -1Question 57 of 57

A 20 cm long string, having a mass of 1.0 g, is fixed at both the ends. The tension in the string is 0.5 N. The string is set into vibrations using an external vibrator of frequency 100 Hz. Find the separation (in cm) between the successive nodes on the string.

Your answer

Answer and worked solution

Official answer: 5

Wave speed. $\mu=\dfrac{10^{-3}}{0.2}=5\times10^{-3}$ kg/m, $v=\sqrt{\dfrac T\mu}=\sqrt{100}=10$ m/s.

Node spacing. The string vibrates at the driving frequency: $\lambda=\dfrac{10}{100}=0.1$ m, so nodes are $\frac\lambda2=5$ cm apart (the 20 cm string holds 4 loops).

Answer: 5

Topics:Standing waves in strings Speed of waves

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IIT-JEE 2009 · Paper 2, Physics question 57